\end{pmatrix}\cdot \begin{pmatrix}
\end{pmatrix}\\
\end{pmatrix}\\
\end{pmatrix}\\
\end{eqnarray*}
Further, $\det(\psi(x+iy))=1,$ as shown in the following computation. Here we use that for $x+iy\in\U{n},$ $|\det(x+iy)|=1.$
\begin{eqnarray*}
\det(\psi(x+iy))&=&\det\begin{pmatrix}
\end{pmatrix}\\
\end{pmatrix}\\
\end{pmatrix}\\
\end{pmatrix}\\
\end{pmatrix}\\
\end{eqnarray*} Thus, $\psi(\U n)\in\SO{2n}.$

Further, $\psi$ is a group homomorphism. For $x+iy,a+ib\in\U n,$ \begin{eqnarray*}
\psi((x+iy)\cdot(a+ib))&=&\begin{pmatrix}
\end{pmatrix}\\
\end{pmatrix}\cdot \begin{pmatrix}
\end{pmatrix}\\
\end{eqnarray*}
Lastly, $\psi$ is injective. Indeed, $\psi(x+iy)=I_{2n}$ if and only if $y=0$ and $x=I_n.$ Hence, $\ker\psi=I_n.$

We conclude that $\psi$ is a group isomorphism of $\U n$ onto its image and we may henceforth identify $\U n$ with $\psi(\U n)\in\SO{2n}.$
\end{proof}
\begin{remark}{\rm [39]}
We will now show that $\SO{2n}/\U{n}$ is indeed a symmetric space. For this, let $$J_n=\begin{pmatrix}
\end{pmatrix}$$ and consider the map $\sigma:\SO{2n}\rightarrow\SO{2n}$ given by $$\sigma(x)=J_nxJ_n^t.$$ If we write $x$ as a block matrix $$x=\begin{pmatrix}
\end{pmatrix},$$ we see that $$\sigma(x)=\begin{pmatrix}
\end{pmatrix}.$$ Hence, if $x\in\U{n}$ then $x$ is fixed by $\sigma.$ Further, we see that $\sigma$ is an ivolutive automorphism since \begin{enumerate}
\item $\sigma(xy)=J_nxyJ_n^t=J_nxJ_n^t J_nyJ_n^t=\sigma(x)\sigma(y),$
\item $\sigma$ is bijective,
\item $\sigma^2(x)=x.$
\end{enumerate}
By Theorem \ref{GKsymspace}, $\SO{2n}/\U{n}$ is a symmetric space.
\end{remark}

Proposition \ref{decompositionso2nun} shows us how to decompose the Lie algebra $\so{2n}$ of $\SO{2n}.$
\begin{proposition}\label{decompositionso2nun}{\rm [39]}
For the orthogonal decomposition $$\so{2n}=\u{n}\oplus\m,$$ where $\u{n}$ is the Lie algebra of $\U{n}\subset\SO{2n},$ a basis of $\m$ is given by $$ \mathcal{B}_\m=\left\{Y_{rs}^{a}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix},Y_{rs}^{b}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix}\ \rvert \ 1\leq r<s\leq n\right\}. $$
\end{proposition}
\begin{proof}
In light of Proposition \ref{embeddingunso2n}, we note that the Lie algebra of $\u{n}$ in $\so{2n}$ is the vector subspace $$\u{n}=\left\{\begin{pmatrix}
\end{pmatrix}\ \rvert \ W^t-W=0,\ \textrm{and} \ Y^t+Y=0\right\}.$$ It can now be easily verified that the subspace $$\m=\left\{\begin{pmatrix}
\end{pmatrix}\in\so{2n} \ \rvert \ Y^t+Y=0,\ \textrm{and} \ W^t+W=0\right\}$$ of $\so{2n}$, spanned by $\mathcal{B}_\m$ as given above, is the orthogonal complement of $\u{n}$ in $\so{2n}.$
\end{proof}

\section{Eigenfunctions on $\SO{2n}/\U n$ }\label{efso2nunsection}
In their paper [22], Gudmundsson, Siffert and Sobak constructed the family of eigenfuntions given in Proposition \ref{so2nunprop}. In Proposition \ref{so2nunnew}, we show that the conditions on the vectors $a,b$ can be relaxed. The proof still works as before. This serves two purposes. We not only obtain more eigenfunctions, but as we will see in Section \ref{cmsso2nun}, the eigenfunctions given in Proposition \ref{so2nunprop} might not always be regular over $0\in\cn$. Further, the computation of the gradient can be very complicated. We  however successfully apply Theorem \ref{GM} to a subset of the newly found eigenfunctions, see Theorem \ref{thmsso2nun}.

\begin{remark}
Notice that for $y\in\U{n},$ $$\Phi(y)=yJ_ny^t=J_n.$$
Hence $\Phi$ is $\U{n}$-invariant, in the sense that for any $y\in\U{n},x\in\SO{2n},$ $$\Phi(xy)=xyJ_ny^tx^t=xJ_nx^t=\Phi(x).$$
\end{remark}

\begin{proposition}{\rm [22]}\label{so2nunprop}
Let $V$ be an isotropic subspace of $\cn^{2n}$ and let $a,b\in V$ be linearly independent. Let $A\in\cn^{2n\times2n}$ be the skew-symmetric matrix $$A=\sum_{j,\alpha=1}^{2n} a_jb_\alpha Y_{j\alpha}.$$ Then $$\hat{\phi}_{a,b}(x)=\trace(A\Phi(x))=\trace(AxJ_nx^t)$$
is a $\U{n}$-invariant eigenfunction on $\SO{2n},$ and hence induces an eigenfunction $\phi_{a,b}$ on the quotient space $\SO{2n}/\U n.$ The eigenvalues $\lambda$ and $\mu$ of $\phi_{a,b}$ satisfy $$\lambda=-2(n-1)\quad \textrm{and}\quad \mu=-1.$$
\end{proposition}
\begin{remark}
Let $X$ be an elements of the Lie algebra $\so{2n}$ of $\SO{2n}.$ Then $X$ acts on the function $\hat{\phi}_{a,b}$ as follows.
\begin{eqnarray*}
X(\hat{\phi}_{a,b})&=&\frac{d}{ds}(\hat{\phi}_{a,b}(x\cdot \exp(sX))\rvert_{s=0}\\
\end{eqnarray*}
We have seen in Remark \ref{decompositionso2nun} that if $X\in\u{n},$ then $X$ is of the form $$X=\begin{pmatrix}
\end{pmatrix},$$ where $W^t-W=0$ and $Y^t+Y=0$.
In this case, $$XJ_n+J_nX^t=\begin{pmatrix}
\end{pmatrix}=0.$$
This demonstrates that $\hat{\phi}_{a,b}$ is $\U{n}$-invariant.
\end{remark}

The following result is the same as Proposition \ref{so2nunprop} from [22], except that the conditions on $a,b$ are slightly weakened.

\begin{proposition}\label{so2nunnew}
Let $a,b\in\cn^{2n}$ be linearly independent vectors satisfying $$(a,a)(b,b)-(a,b)^2=0.$$ Let $A\in\cn^{2n\times2n}$ be the skew-symmetric matrix $$A=\sum_{j,\alpha=1}^{2n} a_jb_\alpha Y_{j\alpha}.$$ Then $$\hat{\phi}_{a,b}(x)=\trace(A\Phi(x))=\trace(AxJ_nx^t)$$
is an $\U{n}$-invariant eigenfunction on $\SO{2n},$ and hence induces an eigenfunction $\phi_{a,b}$ on the quotient space $\SO{2n}/\U n.$
\end{proposition}
\begin{proof}
In this proof we proceed as in the proof of Proposition \ref{so2nunprop}. Let $\Phi:\SO{2n}\rightarrow\SO{2n}$ be the map $$\Phi(x)=x\cdot J_n\cdot x^t.$$ Then, as shown in [22], the components of $\Phi$ satisfy the following:
\begin{eqnarray*}
\tau\left(\Phi_{j\alpha}\right)&=&-2(n-1)\cdot\Phi_{j\alpha},\\
\kappa\left(\Phi_{j\alpha},\Phi_{k\beta}\right)&=&-(\Phi_{j\beta}\Phi_{k\alpha}+\Phi_{jk}\Phi_{\alpha\beta})-(\delta_{k\alpha}\delta_{j\beta}-\delta_{jk}\delta_{\alpha\beta}).
\end{eqnarray*}
Since $A_{j\alpha}=\frac{1}{\sqrt{2}}(a_jb_\alpha-a_\alpha b_j),$ we may now write $$\hat{\phi}_{a,b}(x)=\trace(A\Phi(x))=-\sum_{j,\alpha=1}^{2n} A_{j\alpha}\Phi_{j\alpha}.$$
Using linearity of $\tau,$ we immediately obtain that \begin{eqnarray*}
\tau(\hat{\phi}_{a,b})&=&-\sum_{j,\alpha=1}^{2n} A_{j\alpha}\tau(\Phi_{j\alpha})\\
&=&\sum_{j,\alpha=1}^{2n} A_{j\alpha}\cdot2(n-1)\cdot\Phi_{j\alpha}\\
\end{eqnarray*}
We now calculate $\kappa(\hat{\phi}_{a,b},\hat{\phi}_{a,b}).$
\begin{eqnarray*}
\kappa(\hat{\phi}_{a,b},\hat{\phi}_{a,b})&=&\sum_{j,\alpha,k,\beta=1}^{2n} A_{j\alpha}A_{k\beta}\cdot\kappa(\Phi_{j\alpha},\Phi_{k\beta})\\
&=&-\sum_{j,\alpha,k,\beta=1}^{2n} A_{j\alpha}A_{k\beta}\cdot(\Phi_{j\beta}\Phi_{k\alpha}+\Phi_{jk}\Phi_{\alpha\beta})\\
&&-\sum_{j,\alpha,k,\beta=1}^{2n} A_{j\alpha}A_{k\beta}\cdot(\delta_{k\alpha}\delta_{j\beta}-\delta_{jk}\delta_{\alpha\beta})\\
&=&-\frac{1}{2}\cdot\sum_{j,\alpha,k,\beta=1}^{2n}(a_jb_\alpha-a_\alpha b_j)(a_kb_\beta-a_\beta b_k)\cdot(\Phi_{j\beta}\Phi_{k\alpha}+\Phi_{jk}\Phi_{\alpha\beta})\\
\end{eqnarray*}
The first sum now simplifies to $2\cdot\hat{\phi}_{a,b}^2,$ using the identity $$\hat{\phi}_{a,b}^2=2\cdot\left(\sum_{j,\alpha} a_jb_\alpha\Phi_{j\alpha}\right)^2.$$ Terms of the form $$\left(\sum_{j,\alpha}a_ja_\alpha\Phi_{j\alpha}\right)\cdot\left(\sum_{k,\beta}a_ka_\beta\Phi_{k\beta}\right)$$ appear twice with a plus and twice with a minus sign, and thus they cancel.
By skew-symmetry of $A,$ we have that $$-\sum_{j,k=1}^{2n}A_{jk}A_{kj}+\sum_{j,k=1}^{2n}A_{jk}^2=2\cdot\sum_{j,k=1}^{2n}A_{jk}^2.$$
As in the proof of Proposition \ref{so2nunprop}, we obtain the following equation:
Since we have assumed that $(a,a)(b,b)-(a,b)^2=0$ we conclude that $$\kappa(\hat{\phi}_{a,b},\hat{\phi}_{a,b})=-\hat{\phi}_{a,b}^2.$$
\end{proof}

\section{Compact Minimal Submanifolds of $\SO{2n}/\U n$}\label{cmsso2nun}
We will now investigate the regularity over $0\in\cn$ of the functions given in Propositions \ref{so2nunprop} and \ref{so2nunnew}.

In Example \ref{exampleso2nunold} we show that the functions described in Proposition \ref{so2nunprop} are not necessarily regular over $0\in\cn.$
On the other hand, Theorem \ref{thmsso2nun} shows that a particular subset of the functions given in Proposition \ref{so2nunnew} are indeed regular over $0\in\cn.$ This allows us to construct compact minimal submanifolds using Theorem \ref{GM}.

Remark \ref{remarkso2n} provides us with some useful computations.
\begin{remark}\label{remarkso2n}
Let $\hat{\phi}_{a,b}(x)=\trace(AxJ_nx^t)$ be the function given in Proposition \ref{so2nunprop}.
%Recall that we require both of the linearly independent vectors $a,b\in\cn^{2n}\backslash\{0\}$ to satisfy $(a,a)=(b,b)=0.$
Skew-symmetry of $A$ and $\Phi$ yields \begin{eqnarray*}
\hat{\phi}_{a,b}(x)&=&-\sum_{j,\alpha} A_{j\alpha}\Phi_{j\alpha}\\
&=&-\frac{1}{\sqrt{2}}\cdot\sum_{j,\alpha}\left((a_jb_\alpha-b_ja_\alpha)\cdot\sum_{t=1}^n(-x_{j,n+t}x_{\alpha t}+x_{jt}x_{\alpha,n+t})\right)\\
\end{eqnarray*}
Consider $Y_{rs}^{a}$ and $Y_{rs}^{b}$ as defined in Proposition \ref{decompositionso2nun}. A computation shows that $Y_{rs}^{a}(\hat{\phi}_{a,b})=0$ if and only if %$$0=\sum_{j,\alpha =1}^{2n}(a_jb_\alpha-b_ja_\alpha)\cdot(-x_{\alpha,n+s}\cdot x_{j,r}+x_{\alpha,n+r}\cdot x_{j,s}-x_{\alpha,s}\cdot x_{j,n+r}+x_{\alpha,r}\cdot x_{j,n+s}),$$
and similarly, $Y_{rs}^{b}(\hat{\phi}_{a,b})=0$ if and only if
%$$0=\sum_{j,\alpha =1}^{2n}(a_jb_\alpha-b_ja_\alpha)\cdot(-x_{\alpha,s}\cdot x_{j,r}+x_{\alpha,r}\cdot x_{j,s}+x_{\alpha,n+s}\cdot x_{j,n+r}-x_{\alpha,n+r}\cdot x_{j,n+s}).$$
\end{remark}

\begin{example}\label{exampleso2nunold}
Let $n=2,$ and consider $a=(1,i,0,0),b=(0,0,1,i).$ It is clear that $a,b$ satisfy $(a,a)(b,b)-(a,b)^2$. Let $$x=\begin{pmatrix}
\end{pmatrix}\in\SO{2n}.$$ We will now use the computations from Remark \ref{remarkso2n}. It is easy to see that $$(a,x_1)=(a,x_2)=(b,x_3)=(b,x_4)=0$$ and $$(a,x_3)=i,\ (a,x_4)=-1,\ (b,x_1)=1,\ (b,x_2)=i.$$
Then, since $$(a,x_3)(b,x_1)+(a,x_4)(b,x_2)=i-i=0,$$ it follows that $$\hat{\phi}_{a,b}(x)=0.$$ Further, $Y_{12}^{a}(\hat{\phi}_{a,b})=0 \ \textrm{and}\ Y_{12}^{b}(\hat{\phi}_{a,b})=0.$ Consequently, $\hat{\phi}_{a,b}$ is not regular over $0$ for our choices of $a,b.$
\end{example}
We will now show that with the relaxed conditions of Proposition \ref{so2nunnew}, it is easier to find suitable vectors $a,b$ that render $\hat{\phi}_{a,b}$ regular over $0$. We start with an easy example in the setting $n=2.$
\begin{example}\label{example2so2nun}
Let now $\hat{\phi}_{a,b}$ be as in Proposition \ref{so2nunnew}, with the relaxed condition on $b.$ Let $n=2.$ We set $a=(1,i,0,0),$ and $b=(0,0,1,0).$ Clearly the conditions $$(a,a)=0, (a,b)=0$$ are satisfied, and consequently $\hat{\phi}_{a,b}$ as defined in Proposition \ref{so2nunnew} is an eigenfunction on $\SO{4}/\U{2}.$

We now wish to show that $\hat{\phi}_{a,b}$ is regular over $0.$ For our choice of $a$ and $b$ we have that
Since all entries of $x$ are real-valued, $\hat{\phi}_{a,b}(x)=0$ if and only if the following two equations are satisfied:
\begin{eqnarray}
\end{eqnarray}
Similarly, by separating the real and complex parts, $Y_{12}^{a}(\hat{\phi}_{a,b})=Y_{12}^{b}(\hat{\phi}_{a,b})=0$ if and only if
\begin{eqnarray}
\end{eqnarray}
Lastly, $x\in\SO{4}$ yields the following equations:
\begin{eqnarray}
\end{eqnarray}
We now have 4 equations where entries of the third row and entries of the first appear, namely Equations \ref{so2nunphi3},\ref{so2nunrow1},\ref{so2nunrow2} and \ref{eqso4}. Now assume that for some $x\in\SO{4},$ they all hold true. This means in particular that the non-zero vector $(x_{31},x_{32},x_{33},x_{34})^t$ belongs to the kernel of the matrix $$M=\begin{pmatrix}
\end{pmatrix}.$$
However, $$\det(M)=(x_{11}^2+x_{12}^2+x_{13}^2+x_{14}^2)^2=1,$$
since $x\in\SO{4}.$ By a result from elementary linear algebra, $\ker M=\{0\}.$
This contradicts the assumption that the third row has non-zero entries. It follows that $\hat{\phi}_{a,b}$ is regular over $0.$

As a consequence, we see that $\phi_{a,b}$ is regular over $0$ and that $\phi_{a,b}^{-1}(0)$ is a minimal submanifold, due to Theorem \ref{GM}.
\end{example}
We will now give a more general statement on how to choose vectors $a,b$ such that the corresponding function $\phi_{a,b}$ is regular over $0.$
\begin{theorem}\label{thmsso2nun}
Let  $a,b\in\cn^{2n}$ satisfy the following conditions: \begin{enumerate}
\item $a$ and $b$ are linearly independent,
\item $(a,a)=(a,b)=0,$
\item $b=z\cdot\Tilde{b}$ for some $z\in\cn^*$ and $\Tilde{b}\in\rn^{2n}.$
\end{enumerate}
Let $$A=\sum_{j,\alpha=1}^{2n} a_jb_\alpha Y_{j\alpha}.$$
Then the eigenfunction $$\phi_{a,b}:\SO{2n}/\U{n}\rightarrow\cn,$$ induced by the $\U{n}$-invariant function $\hat{\phi}_{a,b}:\SO{2n}\rightarrow\cn,$ given by $$\hat{\phi}_{a,b}(x)=\trace(AxJ_nx^t),$$ is regular over $0\in\cn$ and further, the fibre $\phi_{a,b}^{-1}(\{0\})$ is a compact minimal submanifold of $\SO{2n}/\U{n}$ of codimension two.
\end{theorem}
\begin{proof}
Certainly $\phi_{a,b}$ satisfies the conditions of Proposition \ref{so2nunnew}, and thus $\phi_{a,b}$ is an eigenfunction on $\SO{2n}/\U{n}.$ Due to Corollary \ref{phi0}, it is clear that $0$ lies in the image of $\phi_{a,b}.$
For $x\in\SO{2n},$ we denote $x_1,\dots,x_{2n}$ the linearly independent columns of $x.$

We will first consider the case $b\in\rn^{2n}.$
Assume towards a contradiction that for a fixed $x\in\SO{2n},$ $\hat{\phi}_{a,b}(x)=0$ and $\hat{\phi}_{a,b}$ is not regular at $x,$ meaning for all $1\leq r<s\leq2n$ we have that $Y_{rs}^{a}(\hat{\phi}_{a,b})=0 \  \textrm{and} \  Y_{rs}^{b}(\hat{\phi}_{a,b})=0.$

In the following we will use the formulae given in Remark \ref{remarkso2n}. Further, we decompose $a$ into its real and imaginary parts $$a=u+i\cdot v,$$ where $u,v\in\rn^{2n}\backslash\{0\}.$
\begin{eqnarray*}
\hat{\phi}_{a,b}(x)
\end{eqnarray*}
Since all entries of the matrix $x$ and the vector $b$ are real-valued, $\hat{\phi}_{a,b}(x)=0$ implies that   \begin{eqnarray*}
\end{eqnarray*}
We now turn our attention to the gradient.
Remark \ref{remarkso2n} states that $Y_{rs}^{a}(\hat{\phi}_{a,b})=0$ if and only if
and similarly, $Y_{rs}^{b}(\hat{\phi}_{a,b})=0$ if and only if

By taking the real parts of the equations $Y_{rs}^{a}(\hat{\phi}_{a,b})=0$ and $Y_{rs}^{b}(\hat{\phi}_{a,b})=0$ for $1\leq r<s\leq2n,$ we obtain the following equations:
\begin{eqnarray*}
0&=&-(u,x_{n+s})  (b,x_{r})+(u,x_{n+r})  (b,x_{s})-(u,x_{s})  (b,x_{n+r})+(u,x_{r})  (b,x_{n+s}),\\
0&=&-(u,x_{s})  (b,x_{r})+(u,x_{r})  (b,x_{s})+(u,x_{n+s})  (b,x_{n+r})-(u,x_{n+r})  (b,x_{n+s}),\\
\end{eqnarray*}
As we have established in Lemma \ref{lemmaiso}, $u\neq0.$  Since the columns $x_1,\dots,x_{2n}$ form an orthonormal basis of $\rn^{2n},$ there exist at least one index $1\leq r\leq2n,$ such that $$(u,x_r)\neq0.$$ Without loss of generality, assume $r=1.$ Further, we note that $$0=(a,b)=(u+iv,b)=(u,b)+i(v,b)=\sum_{t=1}^{2n}(u,x_t)(b,x_t)+i\cdot\sum_{t=1}^{2n}(v,x_t)(b,x_t).$$ Then clearly the real part of the right hand side vanishes, i.e. $$0=\sum_{t=1}^{2n}(u,x_t)(b,x_t).$$
Then for $ 1< k\leq n,$ we obtain the following system of $2(n-1)+2=2n$ equations.
\begin{eqnarray*}
0&=&-(u,x_{k})  (b,x_{r})+(u,x_{r})  (b,x_{k})+(u,x_{n+k})  (b,x_{n+r})-(u,x_{n+r})  (b,x_{n+k}),\\
0&=&-(u,x_{n+k})  (b,x_{r})+(u,x_{n+r})  (b,x_{k})-(u,x_{k})  (b,x_{n+r})+(u,x_{r})  (b,x_{n+k}).
\end{eqnarray*}
These equations are true if and only if the vector $$((b,x_1),\dots,(b,x_{2n}))^t$$ belongs to the kernel of the following matrix $M.$
{\small $$\begin{pmatrix}
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
\end{pmatrix}$$}

It is now easy to see that $M=S+(u,x_1)\cdot I,$ where $S$ is a skew-symmetric real-valued matrix. It is well known that matrices of this form are invertible (as shown in Lemma \ref{SIinvertible}), and consequently $((b,x_1),\dots,(b,x_{2n}))^t=(0,\dots,0).$
We conclude that $b=0$. This contradicts our assumptions, and we conclude that the $\U{n}$-invariant function $\hat{\phi}_{a,b}:\SO{2n}\rightarrow\cn$ is regular over 0. It follows that the compact fibre $\phi_{a,b}^{-1}(0)$ is regular.

According to Theorem \ref{GM}, $\phi_{a,b}^{-1}(\{0\})$ is a compact minimal submanifold of $\SO{2n}/\U n$ of codimension two.

If $b=z\cdot\Tilde{b},$ for some non-zero $z\in\cn$ and $\Tilde{b}\in\rn^{2n},$ then due to linearity of the form $(\cdot,\cdot),$ we may divide all equations by $z$ before taking real and complex parts. We then proceed as in the real case.

\end{proof}

\chapter{The Symmetric Space $\SU{2n}/\Sp{n}$}
We start off with some basic calculations in Section \ref{prelimsu2nspn}, and in Section \ref{sectionefsu2nspn} we look at a family of eigenfunctions, which was constructed by Gudmundsson, Siffert and Sobak in their paper [22]. In Section \ref{cmssu2nspn} we use those eigenfunctions to construct compact minimal submanifolds of the symmetric space $\SU{2n}/\Sp{n}.$

\section{Preliminaries}\label{prelimsu2nspn}
We show how the quaternionic unitary group $\Sp{n}$ can be identified with a subgroup of $\SU{2n}.$ We also present the corresponding orthogonal decomposition of the Lie algebra $\su{n}$ of $\SU{n},$ which we will work with later.
\begin{proposition}\label{spninsu2n}
The map $\psi:\GLH{n}\rightarrow\GLC{2n}$ given by $$\psi:q=z+jw\mapsto\begin{pmatrix}
\end{pmatrix}$$ is an injective homomorphism of Lie groups. Consequently, we may identify $\Sp{n}$ with its image under $\psi.$ Further, $\psi(\Sp{n})\subset\SU{2n}.$
\end{proposition}
\begin{proof}
We will first show that $\psi$ is a homomorphism. \begin{eqnarray*}
\psi((z_1+jw_1)\cdot(z_2+jw_2))&=&\psi(z_1z_2-\Bar{w}_1w_2+j(w_1z_2+\Bar{z}_1w_2))\\
&=&\begin{pmatrix}z_1z_2-\Bar{w}_1w_2& -\Bar{w}_1\Bar{z}_2-{z}_1\Bar{w}_2\\
\end{pmatrix}\\
\end{pmatrix}\cdot \begin{pmatrix}
\end{pmatrix}\\
\end{eqnarray*}
Clearly $\psi$ is injective since $\ker\psi=I_n.$
If $q=z+jw\in\Sp{n},$ then $$q\Bar{q}^t=z\Bar{z}^t+\Bar{w}w^t+j(w\Bar{z}^t-\Bar{z}w^t)=I_n.$$ It follows that
\begin{eqnarray*}
\end{eqnarray*}
Hence, \begin{eqnarray*}
\psi(z+jw)\cdot\overline{\psi(z+jw)}^t&=&\begin{pmatrix}
\end{pmatrix}\\
\end{pmatrix}\\
\end{eqnarray*}
The previous result shows that $|\det(\psi(z+jw))|=1.$ It is left to show that $\det(\psi(z+jw))=1.$ Let $$J_n=\begin{pmatrix}
\end{pmatrix}.$$ As we will show in Lemma \ref{PhiSu2nSpn}, for $z+jw\in\Sp{n},$ it holds that $$\psi(z+jw)\cdot J_n\cdot\psi(z+jw)=J_n.$$  Using an identity of the {\it Pfaffian}, we see that $$\pf(J_n)=\pf\left(\psi(z+jw)\cdot J_n\cdot\psi(z+jw)\right)=\det(\psi(z+jw))\cdot \pf(J_n).$$ Since $\pf(J_n)^2=\det(J_n)\neq 0,$ it follows that $\det(\psi(z+jw))=1.$ A definition of the Pfaffian and detailed statements of its properties can be found in Haber's lecture notes [30].
\end{proof}

\begin{remark}{\rm [39]}
The map $\sigma:\SU{2n}\rightarrow\SU{2n}$ given by $$\sigma(z)=J_n\Bar{z}J_n$$ is an involutive automorphism which fixes $\Sp{n}.$ By Lemma \ref{GKsymspace}, $\SU{2n}/\Sp{n}$ is indeed a symmetric space.
\end{remark}

\begin{proposition}\label{orthdecompsu2nspn}
The Lie algebra $\su{2n}$ of $\SU{2n}$ admits an orthogonal decomposition $$\su{2n}=\sp{n}\oplus\m,$$
where $$\m=\left\{\begin{pmatrix}
\end{pmatrix}\ \rvert \ Z^*+Z=0, \ \trace(Z)=0, \ W^t+W=0\right\}.$$
\end{proposition}

\begin{proof}
In Definition \ref{liealgebraspn}, we stated that the Lie algebra $\sp{n}$ of $\Sp{n}$ can be written as $$\sp{n}=\left\{ \begin{pmatrix}
\end{pmatrix}\in\mathbb{C}^{2n\times2n} \ \rvert \ Z^*+Z=0, \ W^t-W=0 \right\}. $$
Generally, since $X\in\su{2n}$ satisfies $X^*+X=0$ and $\trace(X)=0,$ an element of $\su{2n}$ must be of the form $$X=\begin{pmatrix}
\end{pmatrix},$$ such that $Z^*+Z=Y^*+Y=0,$ and $\trace(Z)+\trace(Y)=0.$
Let  $$\su{2n}=\sp{n}\oplus\m$$ be an orthogonal decomposition. Then  $\m$ has elements of the form $\begin{pmatrix}
\end{pmatrix}, $ where $Z,W$ satisfy \begin{enumerate}
\item $Z^*+Z=0,$
\item $\Im\trace(Z)=0,$
\item $W^t+W=0.$
\end{enumerate}
Since the first condition on $Z$ implies that the real parts of the diagonal entries of $Z$ vanish, we see that due to the second condition $\trace(Z)=0.$
Thus, $$\m=\left\{\begin{pmatrix}
\end{pmatrix}\ \rvert \ Z^*+Z=0, \ \trace(Z)=0, \ W^t+W=0\right\}.$$
\end{proof}

\begin{proposition} Consider the orthogonal decomposition of the Lie algebra $\su{2n}$ of $\SU{2n}$ given by $$\su{2n}=\sp{n}\oplus\m.$$
Then $\m$ is spanned by $$\mathcal{S}_\m=\left\{Y_{rs}^{a}=\frac{1}{2}\begin{pmatrix}
\end{pmatrix},X_{rs}^{a}=\frac{1}{2}\begin{pmatrix}
\end{pmatrix},Y_{rs}^{b}=\frac{1}{2}\begin{pmatrix}
\end{pmatrix},\right.$$ $$\left.Y_{rs}^{c}=\frac{1}{2}\begin{pmatrix}
\end{pmatrix} \  | \  1\leq r<s\leq n\right\}$$
$$\bigcup \ \left\{D_{rs}^{a}=\frac{1}{2}\begin{pmatrix}
\end{pmatrix}, \ \rvert \ 1\leq r<s\leq n\right\},$$ where $D_{rs}=\frac{1}{\sqrt{2}}\cdot(E_{rr}-E_{ss}).$
\end{proposition}
\begin{proof}
In Proposition \ref{orthdecompsu2nspn} we have seen that an element of $\m$ is of the form $$\begin{pmatrix}
\end{pmatrix},$$ where \begin{enumerate}
\item $Z^*+Z=0$,
\item $\trace(Z)=0$ and
\item $ W^t+W=0.$
\end{enumerate}
We must now find generators of all matrices of the forms $Z$ and $W,$ as described above.

We see that such matrices $Z$ must be in the Lie algebra $\su{n}$ of $\SU{n},$ which is is spanned by $$\{Y_{rs},iX_{rs},iD_{rs} \ \rvert \ 1\leq r<s\leq n\}.$$ The corresponding elements in $\mathcal{S}_m$ are $Y_{rs}^{a},X_{rs}^{a},D_{rs}^{a}.$

Further, we note that the subspace of matrices $W\in\cn^{n\times n}$ is generated by $$\{Y_{rs},iY_{rs} \ \rvert \ 1\leq r<s\leq n\}.$$ The corresponding elements in $\mathcal{S}_\m$ are $Y_{rs}^{b}$ and $Y_{rs}^{c}.$
\end{proof}

\section{Eigenfunctions on $\SU{2n}/\Sp{n}$}\label{sectionefsu2nspn}
We now aim to construct eigenfunctions on $\SU{2n}/\Sp{n}.$ Lemma \ref{PhiSu2nSpn} helps us to construct $\Sp{n}$-invariant functions on $\SU{2n}.$ Proposition \ref{propsusp} from Gudmundsson, Siffert and Sobak's paper [22] then provides us with a family of eigenfunctions on $\SU{2n}/\Sp{n}.
\begin{lemma}\label{PhiSu2nSpn}{\rm [31]}

Let $\Phi:\SU{2n}\rightarrow\SU{2n}$ be the map given by $$\Phi(z)=z\cdot J_n\cdot z^t,$$ where $$J_n=\begin{pmatrix}
\end{pmatrix}.$$ An easy computation shows that if $z\in\Sp{n}$ then
$\Phi(z)=J_n$. In particular, $\Phi$ is $\Sp{n}$-invariant.
\end{lemma}

\begin{proposition}{\rm [22]}\label{propsusp}
For non-zero linearly independent elements $a,b\in\mathbb{C}^{2n},$ let $A\in\mathbb{C}^{2n\times2n}$ be the skew-symmetric matrix $$A=\sum_{i,j=1}^{2n} a_ib_jY_{ij}.$$
Let $\Phi$ be the map given by $$\Phi(z)=z\cdot J_n\cdot z^t.$$ Define the function $\hat{\phi}_{a,b}:\SU{2n}\rightarrow\mathbb{C}$ by $$\hat{\phi}_{a,b}(z)=\trace(A\cdot\Phi(z))=\trace(z^tAzJ_n)=-\sum_{j,\alpha=1}^{2n}A_{j\alpha}\Phi_{j\alpha}(z).$$ Then $\hat{\phi}_{a,b}$ is a $\Sp{n}$-invariant eigenfunction on $\SU{2n},$ and thus induces an eigenfunction $\phi_{a,b}$ on the quotient space $\SU{2n}/\Sp{n}.$ The eigenvalues $\lambda$ and $\mu$ of $\phi_{a,b}$ satisfy $$\lambda=-\frac{2(2n^2-n-1)}{n}\quad\textrm{and}\quad\mu=-\frac{2(n-1)}{n}.$$
\end{proposition}

\section{Compact Minimal Submanifolds of $\SU{2n}/\Sp{n}$}\label{cmssu2nspn}
We wish to apply Theorem \ref{GM} to the functions stated in proposition \ref{propsusp}. As Theorem \ref{minsubmsuspn} shows, they are regular over $0\in\cn$ and thus can be used to construct compact minimal submanifolds of $\SU{2n}/\Sp{n}.$

\begin{theorem}\label{minsubmsuspn}
Let $n\geq2$. Let $a,b\in\cn^{2n}$ be linearly independent and let $A$ be the skew-symmetric matrix given by $$A=\sum_{i,j=1}^{2n} a_ib_jY_{ij}.$$ Let $\phi_{a,b}:\SU{2n}/\Sp{n}\rightarrow\cn$  be the eigenfunction induced by $$\hat{\phi}_{a,b}=\trace(z^tAzJ_n).$$ Then $\phi_{a,b}$ is regular over $0\in\cn,$ and $\phi_{a,b}^{-1}(\{0\})$ is a compact minimal submanifold of $\SU{2n}/\Sp{n}$ of codimension two.
\end{theorem}
\begin{proof}
First of all, $\phi_{a,b}$ is an eigenfunction on $\SU{2n}/\Sp{n} $ by Proposition \ref{propsusp}. Due to Corollary \ref{phi0}, $0$ lies in the image of $\phi_{a,b}.$

Assume towards a contradiction that for $z\in\SU{2n}$ we have that $\hat{\phi}_{a,b}(z)=0,$ and for every $Y\in\m,$ $Y(\hat{\phi}_{a,b})=0.$ We will show that under those assumptions, $z$ does not have full rank. We will start by unravelling what our assumptions mean.

For $Y\in\m,$ $$Y(\hat{\phi}_{a,b})=\trace(Az(YJ_n+J_nY^t)z^t).$$ For simplicity we will now write $(a,z_t)=\sum_{j} a_jz_{jt}$ for all $1\leq t\leq 2n.$
It can now be shown that for all $1\leq r<s\leq n,$ $$Y_{rs}^{a}(\hat{\phi}_{a,b})=0, \ Y_{rs}^{b}(\hat{\phi}_{a,b})=0,\ X_{rs}^{a}(\hat{\phi}_{a,b})=0, \ Y_{rs}^{c}(\hat{\phi}_{a,b})=0 \ $$ if and only if
\begin{eqnarray}
\end{eqnarray}
By adding and subtracting Equations \ref{suspeq1} and \ref{suspeq3}, and Equations \ref{suspeq2} and \ref{suspeq4} respectively, we see that for all $1\leq r<s\leq n,$
\begin{eqnarray}
\end{eqnarray}
Since the columns of $\Bar{z}$ form a basis of $\cn^{2n}$, there exists an index $1\leq t\leq 2n$ such that $$\langle a,\Bar{z}_t\rangle =(a,z_t)\neq0.$$ Without loss of generality, $1\leq t\leq n.$ Equations \ref{suspeq5},\ref{suspeq6},\ref{suspeq7} and \ref{suspeq8} now show that for all $1\leq s\leq n, \ s\neq t,$ it holds that $$(b,z_{n+s})=\frac{(a,z_{n+s})}{(a,z_t)}\cdot(b,z_t), \ \textrm{and} \ (b,z_{s})=\frac{(a,z_s)}{(a,z_t)}\cdot(b,z_t).$$
Now note that
$\hat{\phi}_{a,b}(z)=0$ if and only if $$0=(a,z_{n+1})(b,z_{1})+\dots +(a,z_{2n})(b,z_{n})-(a,z_{1})(b,z_{n+1})-\dots-(a,z_{n})(b,z_{2n}).$$
Entering our latest findings into this equation, we note that
\begin{eqnarray*}
0&=&(a,z_{n+t})(b,z_t)+\sum_{s\neq t}(a,z_{n+s})\cdot\frac{(a,z_s)}{(a,z_t)}\cdot(b,z_t)\\
&&-(a,z_t)(b,z_{n+t})-\sum_{s\neq t}(a,z_s)\cdot\frac{(a,z_{n+s})}{(a,z_t)}\cdot(b,z_t)\\
\end{eqnarray*}
Thus, we also have that $$(b,z_{n+t})=\frac{(a,z_{n+t})}{(a,z_t)}\cdot (b,z_{t}).$$
To summarise, we have now established that for every $1\leq s\leq 2n,$ $$\langle b,\Bar{z}_s\rangle =(b,z_{s})=\frac{(a,z_{s})}{(a,z_t)}\cdot(b,z_t)=\frac{(b,z_t)}{(a,z_t)}\cdot\langle a, \Bar{z}_s\rangle.$$
Thus, $$b=\sum_{s=1}^{2n}\langle b,\Bar{z}_s\rangle\cdot \Bar{z}_s=\frac{(b,z_t)}{(a,z_t)}\cdot\sum_{s=1}^{2n}\langle a,\Bar{z}_s\rangle\cdot \Bar{z}_s=\frac{(b,z_t)}{(a,z_t)}\cdot a.$$
This is a contradiction, since we assumed $a$ and $b$ to be linearly independent. It follows that $\phi_{a,b}$ is regular over $0.$
By Theorem \ref{GM}, $\phi_{a,b}^{-1}(\{0\})$ is a minimal submanifold of $\SU{2n}/\Sp{n}.$
\end{proof}

\chapter{The Real Grassmannians $\SO{m+n}/\SO{m}\times\SO{n}$}
Section \ref{efrealgrass} provides us with some useful details on the real Grassmannians. We then study a family of eigenfunctions on $\SO{m+n}/\SO{m}\times\SO{n}$, which stems from Ghandour and Gudmundsson's article [12].
In Section \ref{section2realgrass}, we study minimal submanifolds of $\SO{m+n}/\SO{m}\times\SO{n}$.
We show in Theorem \ref{exsososo2} that none of the aforementioned eigenfunctions is regular over $0.$ Example \ref{exsososo1} illustrates this fact in the setting $m=n=2.$
For $m=3$ and $n=1,$ Example \ref{consolationprize} shows the existence of a compact minimal submanifold.

For the rest of the chapter we assume that the positive integers $m,n$ are not simultaneously equal to $1.$

\section{Eigenfunctions on $\SO{m+n}/\SO{m}\times\SO{n}$}\label{efrealgrass}
We first define the space $\SO{m+n}/\SO{m}\times\SO{n}$ and give a suitable decomposition of the Lie algebra $\so{m+n}$ of $\SO{m+n}.$ Thereafter, we show how to construct eigenfunctions on $\SO{m+n}/\SO{m}\times\SO{n}$, which we will state in Theorem \ref{thsososo}.
\begin{remark}

$\SO{m}\times\SO{n}$ consists of the block matrices of the form $$\SO{m}\times\SO{n}=\left\{ \begin{pmatrix}
\end{pmatrix}\ \rvert \ \ x\in\SO{m}, y\in\SO{n} \right\}. $$
Let $\so{m+n}=\mathfrak{k}\oplus\mathfrak{m}$ be an orthogonal decomposition, where $\mathfrak{k}$ is the Lie algebra of the subgroup $\SO{m}\times\SO{n}$ of $\SO{m+n}$. Then an orthonormal basis for the subspace $\m$ of $\so{m+n}$ is given by
$$\mathcal{B}_\mathfrak{m}=\{ Y_{rs} \ \rvert \ 1\leq r\leq m<s\leq m+n \}. $$
\end{remark}

We now study $\SO{m}\times\SO{n}$-invariant eigenfunctions on $\SO{m+n}.$ They stem from Gudmundsson and Ghandour's work [12]. We start off by looking at the coordinate functions, which are our smallest building blocks.
\begin{lemma}{\rm [12]}\label{lemmasososocoord}
For $1\leq j,k,\alpha,\beta\leq m+n,$ let $x_{j\alpha}:\SO{m+n}\rightarrow\rn$ be the real-valued matrix elements of the standard representation of $\SO{m+n}.$ Then the following relations hold:
$$\tau (x_{j\alpha})=-\frac{m+n-1}{2}\cdot x_{j\alpha},\ \ \textrm{and}\ \ \kappa (x_{j\alpha},x_{k\beta})=-\frac{1}{2}\cdot(x_{j\beta}x_{k\alpha}-\delta_{jk}\delta_{\alpha\beta}).$$
\end{lemma}

\begin{remark}\label{sigmasososo}
Let $$I_{m,n}=\begin{pmatrix}
I_m & 0 \\
0 & -I_n
\end{pmatrix}.$$ A simple calculation shows that if $x\in\SO{m}\times\SO{n}$, then
\begin{eqnarray*}
\sigma(x)=x\cdot I_{m,n}\cdot x^t=I_{m,n}.
\end{eqnarray*} Hence, $\sigma$ is $\SO{m}\times\SO{n}$-invariant. Indeed, for all $w\in\SO{m+n}$ and $x\in\SO{m}\times\SO{n},$ it holds that $$\sigma(wx)=wx\cdot I_{m,n}\cdot x^tw^t=w\cdot I_{m,n}\cdot w^t=\sigma(w).$$
\end{remark}
Using Lemma \ref{lemmasososocoord} and Remark \ref{sigmasososo}, Gudmundsson and Ghandour constructed $\SO{m}\times\SO{n}$-invariant functions on $\SO{m+n}$.
\begin{lemma}{\rm [12]}\label{lemmasososo}
For $1\leq j,\alpha\leq m+n,$ define the function $\hat{\phi}_{j\alpha}:\SO {m+n}\rightarrow\rn$ by $$\hat{\phi}_{j\alpha}(x)=\sum_{t=1}^m x_{jt}x_{\alpha t}.$$ This function is $\SO{m}\times\SO{n}$-invariant and the tension field $\tau $ and the conformality operator $\kappa $ on $\SO{m+n}$ satisfy
\begin{eqnarray*}
\tau (\hat{\phi}_{j\alpha})&=&-(m+n)\hat{\phi}_{j\alpha}+\delta_{j\alpha}\cdot m,\\
\kappa (\hat{\phi}_{j\alpha},\hat{\phi}_{k\beta})&=&-(\hat{\phi}_{j\beta}\cdot\hat{\phi}_{k\alpha}+\hat{\phi}_{jk}\cdot\hat{\phi}_{\alpha\beta})+\frac{1}{2}(\delta_{jk}\hat{\phi}_{\alpha\beta}+\delta_{\alpha\beta}\hat{\phi}_{jk}+\delta_{j\beta}\hat{\phi}_{k\alpha}+\delta_{k\alpha}\hat{\phi}_{j\beta}).
\end{eqnarray*}
\end{lemma}

We now introduce a family of eigenfunctions on the real Grassmannian, which Gudmundsson and Ghandour constructed in [12].
\begin{theorem}{\rm [12]}\label{thsososo}
Let $A$ be a complex symmetric $(m+n)\times(m+n)$ matrix such that $A^2=0.$ Let $$\hat{\phi}_{j\alpha}(x)=\sum_{t=1}^m x_{jt}x_{\alpha t}.$$ Then the $\SO{m}\times\SO{n}$-invariant function $\hat{\Phi}_A: \SO{m+n}\rightarrow\mathbb{C}$ given by $$\hat{\Phi}_A(x)=\sum_{j,\alpha=1}^{m+n} A_{j\alpha}\cdot\hat{\phi}_{j\alpha}(x)$$ induces an eigenfunction $\Phi_A:\SO{m+n}/\SO{m}\times\SO{n}\rightarrow\mathbb{C} $ on the real Grassmannian, if $\rank A=1$ and $\trace A=0.$ In this case,
$$\tau(\Phi_A)=-(m+n)\cdot\Phi_A \ \ \textrm{and} \ \ \kappa(\Phi_A,\Phi_A)=-2\cdot\Phi_A^2.$$
\end{theorem}

\begin{remark}
Lemma \ref{lemmaAaat} implies that all matrices $A$ which satisfy the conditions of Theorem \ref{thsososo} must be of the form $A=aa^t$ for some isotropic $a\in\cn^{m+n}\backslash\{0\}.$ To underline this fact and for consistency of notation, we will write $\Phi_a$ instead of $\Phi_A$ for the remainder of this chapter.
\end{remark}

\section{Minimal Submanifolds of $\SO{m+n}/\SO{m}\times\SO{n}$}\label{section2realgrass}
We now show that none of the functions of the form $\Phi_a,$ as defined in Theorem \ref{thsososo}, are regular over $0\in\cn.$ We will first provide an easy example, and will then prove the general case in Theorem \ref{exsososo2}.

\begin{example}\label{exsososo1}
Let $m=n=2,$ $a=(1,i,0,0)$ and $A=aa^t.$ We now want to find out whether $0$ is a regular value of the function $$\hat{\Phi}_a(x)=\sum_{j,\alpha=1}^{4} a_ja_\alpha\cdot\hat{\phi}_{j\alpha}(x)$$ from Theorem \ref{thsososo}.
Further, for $r=1,2$ and $s=3,4,$ an easy computation shows that $Y_{rs}(\hat{\Phi}_a(x))=0$ is equivalent to
\begin{eqnarray}\label{eqnphiA}
\end{eqnarray}
If this equation holds, then at least one factor must be zero.
Since all entries of $x$ are real-valued, $x_{1t}+ix_{2t}=0$ implies that $x_{1t}=x_{2t}=0.$
Now assume that Equation \ref{eqnphiA} holds for all choices of $r$ and $s.$ Then either $$x_{11}=x_{21}=x_{12}=x_{22}=0\ \  \textrm{or}\ \  x_{13}=x_{23}=x_{14}=x_{24}=0.$$
We further note that $$\hat{\Phi}_a(x)=(x_{11}-x_{22}+i(x_{21}+x_{12}))\cdot(x_{11}+x_{22}+i(x_{21}-x_{12})).$$

Thus, $\nabla\hat{\Phi}_a=0$ at $x$ and $\hat{\Phi}_a(x)=0,$ if $x$ is of the form $$\begin{pmatrix}
\cos\alpha&\sin\alpha&0&0\\
\sin\alpha&-\cos\alpha&0&0\\
\end{pmatrix} $$ for some $\alpha,\theta\in\rn.$
This shows that the set of critical points in the preimage of $\{0\}$ of $\hat{\Phi}_a$ is not discrete. Clearly $0$ is not a regular value.
\end{example}

We will now make a general statement.
\begin{theorem}\label{exsososo2}
Let $a\in\cn\backslash\{0\}$ and assume $(a,a)=0.$ Let
$$\hat{\phi}_{j\alpha}(x)=\sum_{t=1}^m x_{jt}x_{\alpha t}$$ and let the $\SO{m}\times\SO{n}$-invariant function $\hat{\Phi}_a: \SO{m+n}\rightarrow\mathbb{C}$ be given by $$\hat{\Phi}_a(x)=\sum_{j,\alpha=1}^{m+n} a_ja_\alpha\cdot\hat{\phi}_{j\alpha}(x).$$ Define
$\Phi_a$ to be the eigenfunction on $\SO{m+n}/\SO{m}\times\SO{n}$ induced by $\hat{\Phi}_a$. Then $\Phi_a$ is not regular over $0\in\cn.$
\end{theorem}

\begin{proof}
Let $a\in\cn\backslash\{0\}$ be a vector satisfying $(a,a)=0.$ Let $a=u+iv$ be the decomposition of $a$ into its real and imaginary parts.
By Theorem \ref{thsososo} and Lemma \ref{lemmaAaat}, $$\Phi_a:\SO{m+n}/\SO{m}\times\SO{n}\rightarrow\cn$$ is a well defined eigenfunction.
In the following, we will denote by $x_t$ the $t$-th column of a point $x\in\SO{m+n}/\SO{m}\times\SO{n}.$
Through computation it can be shown that \begin{eqnarray*}
\hat{\Phi}_a(x)&=&\sum_{j,\alpha=1}^{m+n} a_ja_\alpha\cdot\hat{\phi}_{j\alpha}(x)\\
&=&\sum_{j,\alpha=1}^{m+n}\sum_{t=1}^m (a_ja_\alpha x_{jt}x_{\alpha t})\\
&=&\sum_{t=1}^{m}\left(\left(\sum_{j=1}^{m+n} a_jx_{jt}\right)\cdot \left(\sum_{\alpha=1}^{m+n} a_\alpha x_{\alpha t}\right)\right)\\
\end{eqnarray*}
Further, for $Y_{rs}\in\mathcal{B}_{\m},$ with $1\leq r\leq m<s\leq m+n,$ we have that $Y_{rs}(\hat{\Phi}_a)=0$ if and only if $$0=\left(\sum_{j=1}^{m+n} a_jx_{jr}\right)\cdot \left(\sum_{\alpha=1}^{m+n} a_\alpha x_{\alpha s}\right)=(a,x_r)\cdot(a,x_s).$$
We will now construct an element $x\in\SO{m+n}$ such that $\hat{\Phi}_a(x)=0$ and for every $Y_{rs}\in\mathcal{B}_\m$, we have that $Y_{rs}(\hat{\Phi}_a)=0$ at $x.$

Assume first that $n\geq2.$ Then we can choose $m$ pairwise orthogonal unit vectors $x_1,\dots,x_m$ of $\rn^{m+n},$ which are also orthogonal to $u$ and $v.$ We may now choose $n$ vectors $x_{m+1},\dots x_{m+n}\in\rn^{m+n}$ that extend
$\{x_1,\dots,x_m\}$ to an orthonormal basis of $\rn^{m+n}.$ Then the matrix $x$ with columns $$(x_1,\dots x_{m+n})$$ clearly belongs to $\O{m+n}.$ If $\det x=-1,$ multiply one of the columns by $-1.$ Hence we may assume that there exists a matrix $x\in\SO{m+n},$ such that the first $m$ columns $x_r, 1\leq r\leq m$ satisfy $$(a,x_r)=(u,x_r)+i(v,x_r)=0.$$  It follows that for all choices $1\leq r\leq m<s\leq m+n,$ we have that $$Y_{rs}(\hat{\Phi}_a)=(a,x_r)\cdot(a,x_s)=0.$$ It trivially holds that $$\hat{\Phi}_a(x)=\sum_{t=1}^{m}(a,x_t)^2=0.$$ We conclude that if $n>1,$ then $0$ is not a regular value of $\hat{\Phi}_a,$ and neither of $\Phi_a.$

Let us now consider the case $m>1, n=1.$ Recall from Lemma \ref{lemmaiso} that $u,v\in\rn^{m+1}$ are linearly independent, non-zero and orthogonal. We now have to adapt our proof, since there do not exist $m$ pairwise orthogonal unit vectors $x_1,\dots,x_m$ of $\rn^{m+1}$ that satisfy %$$\sum_{j=1}^{m+n} (u_j+i\cdot v_j)x_{j,r}=0$$ for all $1\leq r\leq m.$
$$(u+iv, x_r)=0$$ for all $1\leq r\leq m.$
Hence, if $$Y_{r,m+1}(\hat{\Phi}_a)=0$$ for all such $r,$ we require that $$(a,x_{m+1})=0.$$ Let $\Tilde{x}\in\rn^{m+1}$ be any real non-zero unit vector that is orthogonal to both $u$ and $v,$ i.e. it satisfies $$(a,\Tilde{x})=0.$$ Such a vector exists since $m\geq2.$
If $m=2,$ let $x$ be the matrix with columns $$\left(\frac{u}{|u|},\frac{v}{|v|},\Tilde{x}\right).$$ Otherwise,
we may now choose $m-2$ vectors $x_1,\dots,x_{m-2}\in\rn^{m+1}$ that extend $\left\{\frac{u}{|u|},\frac{v}{|v|},\Tilde{x}\right\}$ to an orthonormal basis of $\rn^{m+1}.$
By the same argument as above, we may assume that the matrix $x$ with columns $$\left(\frac{u}{|u|},\frac{v}{|v|},x_1,\dots,x_{m-2},\Tilde{x}\right)$$ belongs to $\SO{m+1}.$
By construction, $$Y_{r,m+1}(\hat{\Phi}_a)=0$$ at $x$ for all $1\leq r\leq m.$ This means that $\hat{\Phi}_a$ is not regular at $x.$
Further, we have that \begin{eqnarray*}
\hat{\Phi}_a(x)&=&\sum_{t=1}^{m}(a,x_t)^2\\
\end{eqnarray*}
Here, we again used Lemma \ref{lemmaiso}. It follows that $\Phi_a$ is not regular over 0.
This proves the statement for $m>1,n=1.$

\end{proof}

As a consequence,the application of Theorem \ref{GM} to functions of the form $\Phi_a$ does not yield compact minimal submanifolds.
For sake of completeness, we will state an Example of a compact minimal submanifold on $\SO{4}/\SO{3},$ namely the famous Clifford Torus.

\begin{example}\label{consolationprize}
(A consolation prize.) As we have seen in Example \ref{spheresoso}, $$S^3\cong\SO{4}/\SO{3}.$$ Since $\SO{1}=\{e\}$ is the trivial group, we may write $$S^3\cong\SO{4}/\SO{3}\times\SO{1}.$$ A well-known minimal submanifold of $S^3$ is the Clifford torus $$T_{{\pi/4}}=\left(\tfrac{1}{\sqrt2}\cdot e^{i\alpha},\tfrac{1}{\sqrt2}\cdot e^{i\beta}\right).$$ We have already discussed it in more detail in Example \ref{clifford}.
\end{example}

\chapter{The Complex Grassmannians $\U{m+n}/\U{m}\times\U{n}$}
We will now turn our attention to the complex Grassmannians.
In Section \ref{sectionefununun} we study eigenfunctions on $\U{m+n}/\U{m}\times\U{n}$.
Lemma \ref{lemma1ununun} corrects a lemma given in Gudmundsson and Ghandour's paper [13]. This allows us to construct a new family of eigenfunctions on $\U{m+n}/\U{m}\times\U{n}$, which is done in Theorem \ref{theigenfunctionununun}.
We then study minimal submanifolds of $\U{m+n}/\U{m}\times\U{n}$ in Section \ref{msununun}.
We show with Theorem \ref{thunununnotregular} that those eigenfunctions are never regular over $0\in\cn.$
Overall, this chapter uses similar arguments as in the last chapter on the real Grassmannians.

\section{Eigenfunctions on $\U{m+n}/\U{m}\times\U{n}$}\label{sectionefununun}
We first show in Lemma \ref{coordfunctionsununun} how the tension field and conformality operator act on the coordinate functions. This Lemma stems from Gudmundsson and Ghandour's paper [13]. We then use it in Lemma \ref{lemma1ununun} to correct a similar statement made in [13]. This allows us to construct a family of eigenfunctions on $\U{m+n}/\U{m}\times\U{n}$, as shown in Theorem \ref{theigenfunctionununun}.

\begin{lemma}{\rm [13]}\label{coordfunctionsununun}
Let $z_{j\alpha}:\U{n}\rightarrow\cn$ be the matrix elements of the standard representation of the unitary group $\U{n}.$ Then, the tension field and conformality operators $\tau,\kappa$ satisfy the following relations:
\begin{eqnarray*}
\tau(z_{j\alpha})&=&-n\cdot z_{j\alpha}, \quad \kappa(z_{j\alpha},z_{k\beta})=-z_{j\beta}z_{k\alpha},\\
\tau(\Bar{z}_{j\alpha})&=&-n\cdot \Bar{z}_{j\alpha}, \quad \kappa(\Bar{z}_{j\alpha},\Bar{z}_{k\beta})=-\Bar{z}_{j\beta}\Bar{z}_{k\alpha},\\
\kappa(z_{j\alpha},\Bar{z}_{k\beta})&=&\delta_{jk}\delta_{\alpha\beta}.\\
\end{eqnarray*}
\end{lemma}
We offer the following correction of Lemma 8.1 of Ghandour and Gudmundsson's paper [13]. We would like to point out that in his thesis [39], Lindström independently came to the same result as us using a different proof.
\begin{lemma}{\rm [13]}\label{lemma1ununun}
For $1\leq j,\alpha\leq m+n,$ we define the complex-valued functions $$\hat{\phi}_{j\alpha}:\U{m+n}\rightarrow\cn$$ on the unitary group by $$\hat{\phi}_{j\alpha}(z)=\sum_{t=1}^m z_{jt}\Bar{z}_{\alpha t}.$$
The tension field $\tau $ and the conformality operator $\kappa $ on the unitary group $\U{m+n}$ satisfy
$$\tau (\hat{\phi}_{j\alpha})=-2(m+n)\cdot\hat{\phi}_{j\alpha}+2m\cdot\delta_{j\alpha},$$
and %$$\kappa (\hat{\phi}_{jk},\hat{\phi}_{lm})=-2\cdot\hat{\phi}_{jk}\hat{\phi}_{lm}.$$
$$\kappa (\hat{\phi}_{j\alpha},\hat{\phi}_{k\beta})=-2\cdot\hat{\phi}_{j\beta}\hat{\phi}_{k\alpha}+\delta_{j\beta}\cdot\hat{\phi}_{k\alpha}+\delta_{k\alpha}\cdot\hat{\phi}_{j\beta}.$$
\end{lemma}
\begin{proof}
Using Lemma \ref{coordfunctionsununun}, we now obtain the following:
\begin{eqnarray*}
\tau(\hat{\phi}_{j\alpha})&=&\sum_{t=1}^m \tau(z_{jt}\Bar{z}_{\alpha t})\\
&=&-2(m+n)\cdot\sum_{t=1}^m z_{jt}\Bar{z}_{\alpha t}+2\cdot\sum_{t=1}^m \delta_{j\alpha}\\
\end{eqnarray*}

\begin{eqnarray*}
\kappa(\hat{\phi}_{j\alpha},\hat{\phi}_{k\beta})&=&\sum_{s=1}^m\sum_{t=1}^m\kappa(z_{js}\Bar{z}_{\alpha s},z_{kt}\Bar{z}_{\beta t})\\
&=&\sum_{s=1}^m\sum_{t=1}^m\Bar{z}_{\alpha s}\Bar{z}_{\beta t}\kappa(z_{js},z_{kt})+z_{js}z_{kt}\kappa(\Bar{z}_{\alpha s},\Bar{z}_{\beta t})\\
&&\quad\quad\quad+\Bar{z}_{\alpha s}z_{kt}\kappa(z_{js},\Bar{z}_{\beta t})+z_{js}\Bar{z}_{\beta t}\kappa(\Bar{z}_{\alpha s},z_{kt})\\
&=&\sum_{s=1}^m\sum_{t=1}^m(-\Bar{z}_{\alpha s}\Bar{z}_{\beta t}z_{jt}z_{ks}+z_{js}z_{kt}\Bar{z}_{\alpha t}\Bar{z}_{\beta s}+\Bar{z}_{\alpha s}z_{kt}\delta_{j\beta}\delta_{st}+z_{js}\Bar{z}_{\beta t}\delta_{\alpha k}\delta_{st})\\
\end{eqnarray*}
\end{proof}

\begin{remark}
For any $z\in\U{m},$ it holds by definition that $z\cdot z^*=I_m.$ Consequently,
\begin{equation*}
(z\cdot z^*)_{j\alpha}=\sum_{t=1}^m z_{jt}\Bar{z}_{\alpha t}=\delta_{j\alpha}.
\end{equation*}
If $j<\alpha,$ it is now clear that the functions $\hat{\phi}_{j\alpha}:\U{m+n}\rightarrow\cn,$ defined by $\hat{\phi}_{j\alpha}(z)=\sum_{t=1}^m z_{jt}\Bar{z}_{\alpha t}$ are $\U{m}\times\U{n}$-invariant and hence induce functions $$\phi_{j\alpha}:\U{m+n}/\U{m}\times\U{n}\rightarrow\cn.$$
\end{remark}

\begin{theorem}{\rm [13], [39]}
For a fixed natural number $1\leq\alpha<m+n,$ the set
$$\mathcal{E}_\alpha=\{\phi_{j\alpha}:\U{m+n}/\U{m}\times\U{n}\rightarrow\cn \ \rvert \ 1\leq j\leq m+n, \ j\neq\alpha\}$$
is an eigenfamily on the complex Grassmannian $\U{m+n}/\U{m}\times\U{n}$ such that the tension field $\tau$ and the conformality operator $\kappa$ satisfy
$$\tau(\phi_{j\alpha})=-2\cdot(m+n)\cdot\phi_{j\alpha},$$ and $$\kappa(\phi_{j\alpha},\phi_{k\alpha})=-2\cdot\phi_{j\alpha}\cdot\phi_{k\alpha}$$ for all $\phi_{j\alpha},\phi_{k\alpha}\in\mathcal{E}_\alpha.$
\end{theorem}
\begin{proof}
The statement  follows immediately from Lemma \ref{lemma1ununun}.
\end{proof}

Building up on the previous lemmas, we now construct a family of $\U{m}\times U{n}$-invariant eigenfunctions on $\U{m+n}$. Those constructions are very similar to the ones we have seen in the last chapter, c.f. Theorem \ref{thsososo}.
\begin{theorem}\label{theigenfunctionununun}
Let $a,b\in\cn^{m+n}\backslash\{0\}$ satisfy $\langle a,\Bar{b}\rangle=0.$ Then the complex-valued function $\hat{\Phi}_{a,b}:\U{m+n}\rightarrow\cn$, given by $$\hat{\Phi}_{a,b}(z)=\sum_{j,\alpha=1}^{m+n} a_jb_\alpha\hat{\phi}_{j\alpha }(z),$$ is an $\U{m}\times\U{n}$-invariant eigenfunction and thus induces an eigenfunction $\Phi_{a,b}:\U{m+n}/\U{m}\times\U{n}\rightarrow\cn.$ In particular,
$$\tau (\hat{\Phi}_{a,b})=-2(m+n)\cdot\hat{\Phi}_{a,b}, \quad \kappa (\hat{\Phi}_{a,b},\hat{\Phi}_{a,b})=-2\cdot\hat{\Phi}_{a,b}^2.$$
\end{theorem}
\begin{proof}
We will make use of Lemma \ref{lemma1ununun}. The rest of the computation is simple.
\begin{eqnarray*}
\tau (\hat{\Phi}_{a,b})&=&\sum_{j,\alpha =1}^{m+n} a_jb_\alpha \tau (\hat{\phi}_{j\alpha })\\
&=&\sum_{j,\alpha =1}^{m+n} a_jb_\alpha (-2(m+n)\cdot\hat{\phi}_{j\alpha }+2m\cdot\delta_{j\alpha })\\
&=&-2(m+n)\cdot\hat{\Phi}_{a,b}+2m\cdot\sum_{j=1}^{m+n} a_jb_j\\
\end{eqnarray*}
In the last step we used that $(a,b)=0.$
\begin{eqnarray*}
\kappa (\hat{\Phi}_{a,b},\hat{\Phi}_{a,b})&=&\sum_{j,\alpha =1}^{m+n}\sum_{k,\beta=1}^{m+n}a_jb_\alpha a_kb_\beta\cdot\kappa (\hat{\phi}_{j\alpha },\hat{\phi}_{k\beta})\\
&=&\sum_{j,\alpha =1}^{m+n}\sum_{k,\beta=1}^{m+n}a_jb_\alpha a_kb_\beta(-2\cdot\hat{\phi}_{j\beta}\hat{\phi}_{k\alpha }+\delta_{j\beta}\hat{\phi}_{k\alpha }+\delta_{k\alpha }\hat{\phi}_{j\beta})\\
&=&-2\cdot\left(\sum_{j,\beta=1}^{m+n} a_jb_\beta\hat{\phi}_{j\beta}\right)\cdot\left(\sum_{k,\alpha =1}^{m+n} a_kb_\alpha \hat{\phi}_{k\alpha }\right)\\
&& +2\cdot\left(\sum_{j=1}^{m+n} a_jb_j\right)\cdot\left(\sum_{k,\alpha =1}^{m+n} a_kb_\alpha \hat{\phi}_{k\alpha }\right)\\
\end{eqnarray*}
\end{proof}

\section{Minimal Submanifolds of $\U{m+n}/\U{m}\times\U{n}$}\label{msununun}
We will now show with Theorem \ref{thunununnotregular} that the functions defined in Theorem \ref{theigenfunctionununun} are not regular over $0\in\cn.$ Therefore we may not apply Theorem \ref{GM}.
We first decompose the Lie algebra $\u{m+n}$ of $\U{m+n},$ which we will use in the proof of Theorem \ref{thunununnotregular}.
\begin{proposition}
Consider the orthogonal decomposition $$\u{m+n}=\mathfrak{k}\oplus \mathfrak{m},$$ where $\k$ is the Lie algebra of $\U{m}\times\U{n},$ and $\m$ is the subspace corresponding to $\U{m+n}/\U{m}\times\U{n}.$
$$\mathcal{B}_\mathfrak{m}=\{Y_{rs}, iX_{rs}\ |\ 1\leq r\leq m,\space m+1\leq s\leq m+n\}$$ is a basis of $\m.$
\end{proposition}
\begin{proof}
As shown in Proposition \ref{liealgebraonun},
$$\mathcal{B}_{\u{m+n}}=\{Y_{rs}, iX_{rs}\ \rvert \ 1\leq r<s\leq m+n\}$$ is a basis of the Lie algebra $\u{m+n}$ of $\U{m+n}.$
It is easily seen that if $1\leq r<s\leq m$ or $m+1\leq r<s\leq m+n,$ then $Y_{rs}$ and $iX_{rs}$ belong to $\k.$ This proves the claim.
\end{proof}
We are now ready to show the main result of this chapter. The statement and proof are similar to Theorem \ref{exsososo2} from the chapter on the real Grassmannians.

\begin{theorem}\label{thunununnotregular}
Let $a,b\in\cn^{m+n}\backslash\{0\}$ with $\langle a,\Bar{b}\rangle =0.$
Let $$\Phi_{a,b}:\U{m+n}/\U{m}\times\U{n}$$ be the eigenfunction induced by the function $\hat{\Phi}_{a,b}:\U{m+n}\rightarrow\cn$ given by $$\hat{\Phi}_{a,b}(z)=\sum_{j,\alpha=1}^{m+n} a_jb_\alpha\hat{\phi}_{j\alpha }(z).$$
Then $\Phi_{a,b}$ is not regular over $0\in\cn.$
\end{theorem}
\begin{proof}
Let $a,b\in\cn^{m+n}\backslash\{0\}$ with $\langle a,\Bar{b}\rangle =0$ be given. %By Theorem \ref{theigenfunctionununun}, the function $\hat{\Phi}_{a,b}:\U{m+n}\rightarrow\cn$ given by $$\hat{\Phi}_{a,b}(z)=\sum_{j,\alpha=1}^{m+n} a_jb_\alpha\hat{\phi}_{j\alpha }(z)$$ induces an eigenfunction $\Phi_{a,b}$ on $\U{m+n}/\U{m}\times\U{n}. $

We now wish to construct $z\in\U{m+n}$ such that for all $X\in\m, \ X(\hat{\Phi}_{a,b})=0$ at $z$ and $\hat{\Phi}_{a,b}(z)=0.$

Let $z_t$ denote the $t$-th column of $z.$
It can be shown that for $1\leq r\leq m<s\leq m+n,$ $Y_{rs}(\hat{\Phi}_{a,b})=0$ if and only if
$$0=\sum_{j,k=1}^{m+n}a_jb_k(z_{js}\Bar{z}_{kr}+z_{jr}\Bar{z}_{ks}),$$ and
$iX_{rs}(\hat{\Phi}_{a,b})=0$ if and only if
Thus, it holds that $Y_{rs}(\hat{\Phi}_{a,b})=iX_{rs}(\hat{\Phi}_{a,b})=0,$ for all $1\leq r\leq m<s\leq m+n$ if and only if \begin{eqnarray}
0&=&\sum_{j,k=1}^{m+n}a_jb_kz_{js}\Bar{z}_{kr}=\langle z_s,\Bar{a}\rangle \langle b,z_r\rangle ,\label{unununyrs}\\
0&=&\sum_{j,k=1}^{m+n}a_jb_kz_{jr}\Bar{z}_{ks}=\langle z_r,\Bar{a}\rangle \langle b,z_s\rangle .\label{unununxrs}
\end{eqnarray}

Similarly, we may write \begin{eqnarray}
\hat{\Phi}_{a,b}(z)&=&\sum_{t=1}^m \langle z_t,\Bar{a}\rangle \langle b,z_t\rangle .\label{eqnphiununun}\end{eqnarray}
Since $\langle \Bar{a},b\rangle =0$ and $a$ and $b$ are non-zero, it is clear that $\Bar{a}$ and $b$ are linearly independent. Further, we can choose $m+n-2$ linearly independent vectors $\{e_1,\dots,e_{m+n-2}\}\in\cn^{m+n}$ such that the set $$S=\{\Bar{a},b,e_1,\dots,e_{m+n-2}\}$$ consists of $m+n$ linearly independent vectors of $\cn^{m+n}.$ Then, for example with the well-known Gram-Schmidt-algorithm, we can derive $m+n$ linearly independent, mutually orthogonal unit vectors of $\cn^{m+n}$ from $S:$
If $n\geq2,$ we let $$z=\left(v_1,\dots,v_{m+n-2},\frac{\Bar{a}}{|\Bar{a}|},\frac{b}{|b|}\right).$$ Then $z\in\U{m+n}$ by construction. Further, for all $1\leq r\leq m, $ we have that $$0=\langle z_r,\Bar{a}\rangle =\langle b,z_r\rangle .$$ It follows that Equations \ref{unununyrs},\ref{unununxrs}, and \ref{eqnphiununun} are satisfied. Hence $z\in\hat{\Phi}_{a,b}^{-1}(\{0\}),$ and $\nabla(\hat{\Phi}_{a,b})=0$ at $z.$

If $n=1,$ let $$z=\left(\frac{\Bar{a}}{|\Bar{a}|},\frac{b}{|b|},v_1,\dots,v_{m+n-2}\right).$$ Again, $z\in\U{m+n}.$ Clearly, Equations \ref{unununyrs} and \ref{unununxrs} are satisfied since $$0=\langle z_n,\Bar{a}\rangle =\langle z_n,b\rangle .$$ Further, $\hat{\Phi}_{a,b}(z)=0$ since $\langle \Bar{a},b\rangle =0.$ In this case we now also have constructed a matrix $z$ such that $z\in\hat{\Phi}_{a,b}^{-1}(\{0\}),$ and $\nabla(\hat{\Phi}_{a,b})=0$ at $z.$

We conclude that $\Phi_{a,b}$ is not regular over $0\in\cn.$
\end{proof}

\chapter{The Quaternionic Grassmannians $\Sp{m+n}/\Sp{m}\times\Sp{n}$}
In Section \ref{efspspspsection} we define eigenfunctions on the quaternionic Grassmannians $G_m(\H^{m+n})=\Sp{m+n}/\Sp{m}\times\Sp{n}$ and show in Section \ref{sectionmsspspsp} that they are not regular over $0$. Our main reference is Ghandour and Gudmundsson's paper [13].

\section{Eigenfunctions on $\Sp{m+n}/\Sp{m}\times\Sp{n}$}\label{efspspspsection}
We now study eigenfunctions on the quaternionic Grassmannians. We first define an eigenfamily found by Gudmundsson and Ghandour in their paper [13]. In Theorem \ref{thspspspnew}, we then define a new eigenfunction, which can be seen as the analogue to the functions given in Theorems \ref{thsososo} and \ref{theigenfunctionununun}.
\begin{lemma}{\rm [13]}\label{quatgrasscoord}
For $1\leq j,\alpha\leq n,$ let $z_{j\alpha},w_{j\alpha}:\Sp{n}\rightarrow\cn$ be the complex-valued matrix elements of the standard representation of the quaternionic unitary group $\Sp{n}$. Then, the tension field $\tau$ and the conformality operator $\kappa$ on $\Sp{n}$ satisfy the following relations:
$$\tau(z_{j\alpha})=-\frac{2n+1}{2}\cdot z_{j\alpha}, \ \tau(w_{j\alpha})=-\frac{2n+1}{2}\cdot w_{j\alpha},$$
$$\kappa(z_{j\alpha},z_{k\beta})=-\frac{1}{2}\cdot z_{j\beta}z_{k\alpha}, \ \kappa(w_{j\alpha},w_{k\beta})=-\frac{1}{2}\cdot w_{j\beta}w_{k\alpha},$$
$$\tau(\overline{z}_{j\alpha})=-\frac{2n+1}{2}\cdot \overline{z}_{j\alpha}, \ \tau(\overline{w}_{j\alpha})=-\frac{2n+1}{2}\cdot \overline{w}_{j\alpha},$$
$$\kappa(\overline{z}_{j\alpha},\overline{z}_{k\beta})=-\frac{1}{2}\cdot \overline{z}_{j\beta}\overline{z}_{k\alpha}, \ \kappa(\overline{w}_{j\alpha},\overline{w}_{k\beta})=-\frac{1}{2}\cdot \overline{w}_{j\beta}\overline{w}_{k\alpha},$$
$$\kappa(\overline{z}_{j\alpha},\overline{w}_{k\beta})=-\frac{1}{2}\cdot \overline{w}_{j\beta}\overline{z}_{k\alpha},\  \kappa({z}_{j\alpha},\overline{z}_{k\beta})=\frac{1}{2}\cdot( w_{j\beta}\overline{w}_{k\alpha}+\delta_{jk}\delta_{\alpha\beta}),$$
$$ \kappa({z}_{j\alpha},\overline{w}_{k\beta})=-\frac{1}{2}\cdot w_{j\beta}\overline{w}_{k\alpha},\ \kappa({w}_{j\alpha},\overline{z}_{k\beta})=-\frac{1}{2}\cdot w_{j\beta}\overline{z}_{k\alpha},$$
$$\kappa({w}_{j\alpha},\overline{w}_{k\beta})=\frac{1}{2}\cdot( z_{j\beta}\overline{z}_{k\alpha}+\delta_{jk}\delta_{\alpha\beta}),\ \kappa(z_{j\alpha},w_{k\beta})=-\frac{1}{2}\cdot w_{j\beta}z_{k\alpha}.$$

\end{lemma}
From Lemma \ref{quatgrasscoord}, Ghandour and Gudmundsson obtained the following result.
\begin{lemma}{\rm [13]}\label{efspspsp}
For $1\leq j<\alpha\leq m+n$ we define the complex-valued functions $\hat{\phi}_{j\alpha}:\Sp{m+n}\rightarrow\cn$ by $$\hat{\phi}_{j\alpha}(q)=\sum_{t=1}^m(z_{jt}\Bar{z}_{\alpha t}+w_{jt}\Bar{w}_{\alpha t}).$$ Then the tension field $\tau $ and the conformality operator $\kappa $ on $\Sp{m+n}$ satisfy $$\tau (\hat{\phi}_{j\alpha})=-2(m+n)\cdot\hat{\phi}_{j\alpha}, \quad \kappa (\hat{\phi}_{j\alpha},\hat{\phi}_{k\alpha})=-\hat{\phi}_{j\alpha}\hat{\phi}_{k\alpha}, $$ where $j,k\neq\alpha.$
\end{lemma}

\begin{remark}
The functions $\hat{\phi}_{j\alpha}$ are all $\Sp{m}\times\Sp{n}$-invariant and hence induce functions $\phi_{j\alpha}:\Sp{m+n}/\Sp{m}\times\Sp{n}\rightarrow\cn$ on the quotient space.
\end{remark}
Lemma \ref{efspspsp} led Ghandour and Gudmundsson to an eigenfamily on the quaternionic Grassmannians.
\begin{theorem}{\rm [13]}
For a fixed natural number $1\leq r<m+n,$ the set
$$\mathcal{E}_r=\{\phi_{j\alpha}:\Sp{m+n}/\Sp{m}\times\Sp{n}\rightarrow\cn \ \rvert \ 1\leq j\leq r<\alpha\leq m+n\}$$ is an eigenfamily on the quaternionic Grassmannian $\Sp{m+n}/\Sp{m}\times\Sp{n}$ such that the tension field $\tau$ and conformality operator $\kappa$ satisfy
$$\tau(\phi)=-2(m+n)\cdot\phi,\quad \kappa(\phi,\psi)=-\phi\cdot\psi$$ for all $\phi,\psi\in\mathcal{E}_r.$
\end{theorem}
In the following, we will generalise Lemma \ref{efspspsp}. Here, we eased our restrictions on the indices $j$ and $\alpha.$
\begin{lemma}\label{quatgrassphigeneral}
For $1\leq j,\alpha,k,\beta\leq m+n,$ we define the complex-valued functions $\hat{\phi}_{j\alpha}:\Sp{m+n}\rightarrow\cn$ by $$\hat{\phi}_{j\alpha}(q)=\sum_{t=1}^m(z_{jt}\Bar{z}_{\alpha t}+w_{jt}\Bar{w}_{\alpha t}).$$ Then, the tension field $\tau $ and the conformality operator $\kappa $ on $\Sp{m+n}$ satisfy $$\tau (\hat{\phi}_{j\alpha})=-2(m+n)\cdot\hat{\phi}_{j\alpha}+2\cdot\delta_{j\alpha}, \quad \kappa (\hat{\phi}_{j\alpha},\hat{\phi}_{k\beta})=-\hat{\phi}_{j\beta}\hat{\phi}_{k\alpha}+\frac{1}{2}\cdot\hat{\phi}_{k\alpha}\delta_{j\beta}+\frac{1}{2}\cdot\hat{\phi}_{j\beta}\delta_{k\alpha}. $$
\end{lemma}
\begin{proof}
The result follows from a computation employing Lemma \ref{quatgrasscoord}.
\end{proof}
The following Theorem is inspired by Theorems \ref{thsososo} and \ref{theigenfunctionununun} in the settings of the real and complex Grassmannians respectively.
\begin{theorem}\label{thspspspnew}
Let $a\in\cn^{m+n}\backslash\{0\}$ satisfy $(a,a)=0.$ Then the complex-valued function $\hat{\Phi}_a:\Sp{m+n}\rightarrow\cn$ given by
$$\hat{\Phi}_a=\sum_{j,\alpha}a_ja_\alpha\hat{\phi}_{j\alpha}$$ satisfies $$\tau (\hat{\Phi}_a)=-2(m+n)\cdot\hat{\Phi}_a\ \textrm{and}\ \kappa (\hat{\Phi}_a,\hat{\Phi}_a)=-\hat{\Phi}_a^2.$$ Since $\hat{\Phi}_a$ is $\Sp{m}\times\Sp{n}$-invariant, it induces an eigenfunction $\Phi_a$ on the quaternionic Grassmannian $\Sp{m+n}/\Sp{m}\times\Sp{n}.$
\end{theorem}
\begin{proof}
We will make use of Lemma \ref{quatgrassphigeneral}. For the tension field, we obtain the following computation.
\begin{eqnarray*}
\tau (\hat{\Phi}_a)&=&\tau \left(\sum_{j,\alpha}a_ja_\alpha\hat{\phi}_{j\alpha}\right)\\
&=&\sum_{j,\alpha}a_ja_\alpha\tau (\hat{\phi}_{j\alpha})\\
&=&-2(m+n)\hat{\Phi}_a+2\cdot\sum_j a_j^2\\
\end{eqnarray*}
Lastly, we compute the conformality operator.
\begin{eqnarray*}
\kappa (\hat{\Phi}_a,\hat{\Phi}_a)&=&\kappa \left(\sum_{j,\alpha}a_ja_\alpha\hat{\phi}_{j\alpha},\sum_{k,\beta}a_ka_\beta\hat{\phi}_{k\beta}\right)\\
&=&\sum_{j,\alpha,k,\beta}a_ja_\alpha a_ka_\beta\cdot\kappa (\hat{\phi}_{j\alpha},\hat{\phi}_{k\beta})\\
&=&\sum_{j,\alpha,k,\beta}a_ja_\alpha a_ka_\beta\cdot\left(-\hat{\phi}_{j\beta}\hat{\phi}_{k\alpha}+\frac{1}{2}\cdot\hat{\phi}_{k\alpha}\delta_{j\beta}+\frac{1}{2}\cdot\hat{\phi}_{j\beta}\delta_{k\alpha}\right)\\
&=&-\hat{\Phi}_a^2+\sum_{j,\alpha,k,\beta}a_ja_\alpha a_ka_\beta\cdot\left(\frac{1}{2}\cdot\hat{\phi}_{k\alpha}\delta_{j\beta}+\frac{1}{2}\cdot\hat{\phi}_{j\beta}\delta_{k\alpha}\right)\\
\end{eqnarray*}
\end{proof}

\section{Minimal Submanifolds of $\Sp{m+n}/\Sp{m}\times\Sp{n}$}\label{sectionmsspspsp}
We first define a basis of the tangent space of $\Sp{m+n}/\Sp{m}\times\Sp{n}$. We use this to show that the eigenfunctions given in Section \ref{efspspspsection} are not regular.
\begin{proposition}\label{liealgspspsp}
For $G=\Sp{m+n},$ and $K=\Sp{m}\times\Sp{n},$ we have the orthogonal decomposition $\g=\k\oplus\m,$ where $\k$ is the Lie algebra of $K$ and $\m$ is the subspace associated with $M=\Sp{m+n}/\Sp{m}\times\Sp{n}.$ It follows that an
orthonormal basis of $\m$ is given by
\begin{eqnarray*}
\mathcal{B}_{\m}
\end{pmatrix}, \ X_{rs}^{a}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix},\right.\\
&&\ \ X_{rs}^{b}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix},   X_{rs}^{c}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix}\\
&& \ \ \rvert \ 1\leq r\leq m<s\leq m+n\Big\}.
\end{eqnarray*}
\end{proposition}
\begin{proof}
In Proposition \ref{liealgebraspn}, we have seen that a basis of the Lie algebra $\sp{m+n}$ of $\Sp{m+n}$ is given by \begin{eqnarray*}
\mathcal{B}_{\sp{m+n}}&=&\left\{Y_{rs}^{a}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix}, \ X_{rs}^{a}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix},\right. \\
&&\ \ X_{rs}^{b}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix}, X_{rs}^{c}=\frac{1}{\sqrt{2}}\begin{pmatrix}
\end{pmatrix} \ \\
&&\ \ \rvert \ 1\leq r<s\leq m+n\Big\}.
\end{eqnarray*}
It is clear that if $1\leq r<s\leq m$ or $m+1\leq r<s\leq m+n,$ then the elements $Y_{rs}^{a},X_{rs}^{a},X_{rs}^{b},X_{rs}^{c}$ belong to $\mathcal{B}_\k.$ The fact that $\mathcal{B}_{\sp{m+n}}$ is the disjoint union of $\mathcal{B}_\k$ and $\mathcal{B}_\m$ now proves the statement.
\end{proof}

In the following, we will show that the eigenfunctions defined on Lemma \ref{efspspsp} are not regular over $0\in\cn.$ In the proof, we will proceed similar to Theorems \ref{exsososo2} and \ref{thunununnotregular}.
\begin{theorem}
Let $1\leq j<\alpha\leq m+n.$ Let $\phi_{j\alpha}:\Sp{m+n}/\Sp{m}\times\Sp{n}\rightarrow\cn$ be the eigenfunction on the quaternionic Grassmannians induced by the function $\hat{\phi}_{j\alpha}:\Sp{m+n}\rightarrow\cn$ defined by $$\hat{\phi}_{j\alpha}(q)=\sum_{t=1}^m(z_{jt}\Bar{z}_{\alpha t}+w_{jt}\Bar{w}_{\alpha t}).$$ Then $\phi_{j\alpha}$ is not regular over $0\in\cn.$
\end{theorem}
\begin{proof}
We know from Lemma \ref{efspspsp} that $\phi_{j\alpha}:\Sp{m+n}/\Sp{m}\times\Sp{n}\rightarrow\cn$ is indeed an eigenfunction, which justifies the statement. We wish to construct an element $q=z+jw\in\Sp{m+n}$ such that $\hat{\phi}_{j\alpha}(q)=0$ and for every $X\in\mathcal{B}_\m$ as specified in Proposition \ref{liealgspspsp}, we have that $X(\hat{\phi}_{j\alpha})=0$ at $q.$
A computation shows that for $1\leq r\leq m<s\leq m+n,$ the following holds.
\begin{eqnarray*}
Y_{rs}^{a}(\hat{\phi}_{j\alpha})&=&\frac{-1}{\sqrt{2}}(z_{js}\Bar{z}_{\alpha r}+z_{jr}\Bar{z}_{\alpha s}+w_{js}\Bar{w}_{\alpha r}+w_{jr}\Bar{w}_{\alpha s}),\\
X_{rs}^{a}(\hat{\phi}_{j\alpha})&=&\frac{i}{\sqrt{2}}(z_{js}\Bar{z}_{\alpha r}-z_{jr}\Bar{z}_{\alpha s}-w_{js}\Bar{w}_{\alpha r}+w_{jr}\Bar{w}_{\alpha s}),\\
X_{rs}^{b}(\hat{\phi}_{j\alpha})&=&\frac{i}{\sqrt{2}}(w_{js}\Bar{z}_{\alpha r}-z_{jr}\Bar{w}_{\alpha s}+z_{js}\Bar{w}_{\alpha r}-w_{jr}\Bar{z}_{\alpha s}),\\
X_{rs}^{c}(\hat{\phi}_{j\alpha})&=&\frac{1}{\sqrt{2}}(-w_{js}\Bar{z}_{\alpha r}-z_{jr}\Bar{w}_{\alpha s}+z_{js}\Bar{w}_{\alpha r}+w_{jr}\Bar{z}_{\alpha s}).
\end{eqnarray*}
Hence $$Y_{rs}^{a}(\hat{\phi}_{j\alpha})=0, \ X_{rs}^{a}(\hat{\phi}_{j\alpha})=0,\ X_{rs}^{b}(\hat{\phi}_{j\alpha})=0, \ \textrm{and}\ X_{rs}^{c}(\hat{\phi}_{j\alpha})=0$$ if and only if \begin{eqnarray}
0&=&z_{js}\Bar{z}_{\alpha r}+w_{jr}\Bar{w}_{\alpha  s},\label{spspspeqns1}\\
0&=&z_{jr}\Bar{z}_{\alpha s}+w_{js}\Bar{w}_{\alpha r},\label{spspspeqns2}\\
0&=&w_{js}\Bar{z}_{\alpha r}-w_{jr}\Bar{z}_{\alpha s},\label{spspspeqns3}\\
0&=&z_{jr}\Bar{w}_{\alpha s}-z_{js}\Bar{w}_{\alpha r}.\label{spspspeqns4}
\end{eqnarray}
Let $e_j,e_\alpha$ be the $j$-th and $\alpha$-th standard unit vectors of $\rn^{m+n}$ respectively. Let $v_1,\dots,v_{m+n-2}$ denote $m+n-2$ pairwise orthogonal unit vectors of $\cn^{m+n}$, which are also orthogonal to $e_j,e_\alpha.$ Such vectors exist, for example we may just take the remaining elements of the canonical basis above.

If $n\geq2,$ let $z=(v_1,\dots,v_{m+n-2},e_j,e_\alpha)$ and $w=0.$ By construction, $q=z+jw$ satisfies $$I_n=z\Bar{z}^t+\Bar{w}w^t, \quad 0=w\Bar{z}^t-\Bar{z}w^t.$$ Thus $q\in\Sp{m+n}.$ For every column $z_t$ of $z,$ we have that $$z_{jt}=\langle z_t,e_j\rangle $$ and $$\Bar{z}_{\alpha t}=\langle e_\alpha,z_t\rangle .$$ Since $v_1,\dots,v_{m+n-2}$ were chosen to satisfy $$\langle v_t,e_j\rangle =\langle e_\alpha,v_t\rangle =0$$ for every $1\leq t\leq m+n-2,$ we see that \begin{eqnarray*}
\hat{\phi}_{j\alpha}(z+jw)&=&\sum_{t=1}^m(z_{jt}\Bar{z}_{\alpha t}+w_{jt}\Bar{w}_{\alpha t})\\
&=&\sum_{t=1}^m\langle z_t,e_j\rangle \langle e_\alpha,z_t\rangle \\
\end{eqnarray*}
Thus, $\hat{\phi}_{j\alpha}(q)=0.$
Further, for every $1\leq r\leq m, $ clearly $z_{jr}=\langle z_r,e_j\rangle =0$ and $\Bar{z}_{\alpha r}=\langle e_\alpha,z_r\rangle =0.$ Since all entries of $w$ vanish, Equations \ref{spspspeqns1} through \ref{spspspeqns4} are satisfied.
We conclude that $\hat{\phi}_{j\alpha}$ is not regular at $q$.
This concludes the proof for $n\geq2.$

In the case that $n=1,$ we let $z=(e_j,e_\alpha,v_1,\dots,v_{m+n-2})$ and $w=0.$ A similar computation as above shows that $\hat{\phi}_{j\alpha}$ is not regular at $z+jw,$ and $\hat{\phi}_{j\alpha}(z+jw)=0$.

\end{proof}

\appendix
\chapter{Notation and Useful Lemmas}

\begin{remark}
We denote $$(\cdot,\cdot):\cn^n\times\cn^n\rightarrow\cn,$$to be the bilinear form $$(x,y)=\sum_{k=1}^n x_ky_k.$$ When restricted to $\rn^n\times\rn^n,$ this yields the standard inner product on $\rn^n.$ We call a vector $a\in\cn^n$ isotropic if it satisfies $(a,a)=0.$

The standard inner product on $\cn^n$ $$\langle \cdot,\cdot\rangle :\cn^n\times\cn^n\rightarrow\cn$$ is given by $$\langle z,w\rangle =\sum_{k=1}^n z_k\Bar{w}_k.$$
\end{remark}
The following statement is widely known and the proof is a standard exercise in Linear Algebra.

\begin{lemma}\label{lemmaiso}
Let $$(\cdot,\cdot):\cn^n\times\cn^n\rightarrow\cn,$$ be the bilinear form $$(x,y)=\sum_{k=1}^n x_ky_k.$$
Let $a=u+i\cdot v\in\cn^n$ be a non-zero vector, where $u,v\in\rn^n.$ If $(a,a)=0,$ then $u,v$ are both non-zero, orthogonal and linearly independent. Further, $$|u|^2=|v|^2.$$
\end{lemma}
\begin{proof}
Let $a\in\cn\backslash\{0\}$ and assume $(a,a)=0.$ Note that
$$0=(a,a)=(u,u)+2i(u,v)-(v,v).$$ It follows that $|u|^2=|v|^2$ and $(u,v)=0.$ Since $a\neq0,$ at least one of $u,v$ is non-zero. But from the previous assertion it follows that both $u,v$ are non-zero.
If $u,v$ were linearly dependent, then $v=\lambda\cdot u$ for some $\lambda\in\rn\backslash\{0\}.$ But then $$0=(u,v)=(u,\lambda\cdot u)=\lambda\cdot(u,u)$$ implies that both $u=v=0,$ since $(\cdot,\cdot)$ defines the standard inner product on $\rn^n\times\rn^n.$
Thus, $u$ and $v$ must be linearly independent.
\end{proof}
The following lemma was used in the proof of Theorem \ref{thmsso2nun} and is covered in basic level courses in Linear Algebra or Matrix Theory.
\begin{lemma}\label{SIinvertible}
Suppose that $S$ is a skew-symmetric and real-valued $n\times n$ matrix. Then $S+I$ is invertible.
\end{lemma}
\begin{proof}
Assume towards a contradiction that $S+I$ is not invertible. Then there exists a non-zero vector $x\in\rn^n$ such that $$(S+I)\cdot x=S\cdot x+x=0.$$ Equivalently, $$S\cdot x=-x.$$ Consider the following computation.
\begin{eqnarray*}
\end{eqnarray*}
Hence $x=0,$ which is a contradiction. We conclude that $S+I$ must be invertible.
\end{proof}

The next statement was used in the chapter on the real Grassmannians. This is well known and is partially discussed in Gudmundsson and Ghandour's work [12].

\begin{lemma}\label{lemmaAaat}
Let $A\in\cn^{n\times n}\backslash\{0\}$ be a symmetric matrix. Then $\rank A=1$ if and only if $A=aa^t$ for some $a\in\cn^n\backslash\{0\}.$ In this case, $\trace A=0$ if and only if $a$ is isotropic, and further, $A^2=0$.
\end{lemma}
\begin{proof}
Let $a\in\cn^n$ and $A=aa^t.$ Since any two rows of $A$ are multiples of one another, the rank of $A$ is one.

Conversely, assume $A$ has rank one. Then the columns are all multiples of some $v\in\cn^n.$ Hence for the standard basis $e_1,\dots,e_n,$ $$A\cdot e_i=\lambda_i\cdot v.$$ But this means that if we write $w=(\lambda_1,\dots\lambda_n)^t,$ $A=vw^t.$ It can now be shown that due to symmetry, $v=w.$

The second statement follows immediately from the fact that $$(a,a)=a_1^2+\dots+a_{m+n}^2=\trace(A).$$
If $(a,a)=0$, then for all $1\leq i,j\leq n,$ $$(A^2)_{ij}=\sum_{t=1}^n A_{it}A_{tj}=a_ia_j\sum_{t=1}^n a_t^2=0.$$ This shows that $A^2=0.$
\end{proof}

%\chapter{Computations in Maple}

%    \adjustimage{scale=1,left}{Maple-1.eps}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  %%%%%%%%%%%%%%%%%%%%%%%

\begin{thebibliography}{99}\label{biblio}
\normalsize
[1]
N. Agrawal, D. Khattar,
{\it Group Theory}, Springer (2023).

[2]
A. Arvanitoyeorgos,
{\it An Introduction to Lie Groups and the Geometry of Homogeneous Spaces}, Student mathematical library {\bf 22}, American Mathematical Society (2003).

[3]
P. Baird, J. Eells,
{\it A conservation law for harmonic maps},
Geometry Symposium Utrecht 1980, Lecture Notes in Mathematics {\bf 894}, 1-25, Springer (1981).

[4]
P. Baird, S. Gudmundsson,
{\it $p$-hamonic maps and minimal submanifolds},
Math. Ann. {\bf 294} (1992), 611-624.

[5]
P.~Baird, J.~C. Wood,
{\it Harmonic morphisms between Riemannian manifolds},
London Math. Soc. Monogr. No. {\bf 29}
Oxford Univ. Press (2003).

[6]
S. Brendle, {\it Embedded minimal tori in $S^3$ and the Lawson conjecture}, Acta Math. {\bf 211} (2013), 177-190.

[7]
F.E. Burstall, {\it Harmonic Riemannian submersions between Riemannian symmetric spaces of noncompact type}, (preprint), 2024.

[8]
J. Eells, J.H. Sampson, {\it Harmonic mappings of Riemannian manifolds}, Amer. J. Math. {\bf 86} (1964), 109-160.

[9]
J.H. Eschenburg, {\it Lecture Notes on Symmetric Spaces}, University of Augsburg,\\ \href{https://myweb.rz.uni-augsburg.de/~eschenbu/symspace.pdf}{\tt www.myweb.rz.uni-augsburg.de/~eschenbu/symspace.pdf}

[10]
J.M. Gegenfurtner,
{\it CMC Tori in the Generalised
Berger Spheres and their Duals}, Bachelor's Thesis, Lund University (2022),\\
\href{https://arxiv.org/abs/2207.03535}{\tt www.arxiv.org/abs/2207.03535}

[11]
E. Ghandour, S. Gudmundsson,
{\it Complex-valued $(p,q)$-harmonic morphisms from Riemannian manifolds},
J. Geom. Anal. {\bf 31} (2021), 11386-11409.

[12]
E. Ghandour, S. Gudmundsson,
{\it Explicit $p$-harmonic functions on the real Grassmannians},
Adv. Geom. {\bf 23} (2023), 315–321.

[13]
E. Ghandour, S. Gudmundsson,
{\it Explicit harmonic morphisms and $p$-harmonic functions from the complex and quaternionic Grassmannians},
Ann. Global Anal. Geom. {\bf 64} 15 (2023).

[14]
{\it The Bibliography of Harmonic Morphisms},\\ \href{http://www.matematik.lu.se/matematiklu/personal/sigma/harmonic/bibliography.html}
{\tt www.matematik.lu.se/ matematiklu/personal/sigma/harmonic/bibliography.html}

[15]
S. Gudmundsson,
{\it An Introduction to Riemannian Geometry},  Lecture Notes in Mathematics,
Lund University (2022),\\
\href{http://www.matematik.lu.se/matematiklu/personal/sigma/Riemann.pdf}{\tt www.matematik.lu.se/matematiklu/personal/sigma/Riemann.pdf}

[16]
S. Gudmundsson, S. Montaldo, A. Ratto,
{\it Biharmonic functions on the classical compact simple Lie groups},
J. Geom. Anal. {\bf 28} (2018), 1525-1547.

[17]
S. Gudmundsson, T. J. Munn,
{\it Minimal submanifolds via complex-valued eigenfunctions},
J. Geom. Anal. {\bf 34} 190 (2024).

[18]
S. Gudmundsson, T. J. Munn,
{\it Harmonic morphisms on Lie groups and minimal conformal foliations of codimension two},
J. Geom. Phys. {\bf 198} (2024).

[19]
S. Gudmundsson, A. Sakovich,
{\it Harmonic morphisms from the classical compact semisimple Lie groups},
Ann. Global Anal. Geom. {\bf 33} (2008), 343-356.

[20]
S. Gudmundsson, A. Sakovich,
{\it Harmonic morphisms from the classical non-compact semisimple Lie groups},
Differential Geom. Appl. {\bf 27} (2009), 47-63.

[21]
S. Gudmundsson, A. Siffert,
{\it New biharmonic functions on the compact Lie groups $\SO{n},$ $\SU{n},$ $\Sp{n}$},
J. Geom. Anal. {\bf 31} (2021).

[22]
S. Gudmundsson, A. Siffert, M. Sobak,
{\it Explicit proper $p$-harmonic functions on the Riemannian symmetric spaces $\SU n/\SO n$, $\Sp n/\U n$, $\SO {2n}/\U n$, $\SU{2n}/\Sp n$},
J. Geom. Anal. {\bf 32} 147 (2022).

[23]
S. Gudmundsson, M. Sobak,
{\it Proper $r$-harmonic functions from Riemannian manifolds},
Ann. Global Anal. Geom. {\bf 57} (2020), 217-223.

[24]
S. Gudmundsson, M. Svensson,
{\it Harmonic morphisms from the Grassmannians and their non-compact duals},
Ann. Global Anal. Geom. {\bf 30} (2006), 313-333.

[25]
S. Gudmundsson, M. Svensson,
{\it Harmonic morphisms from the compact semisimple Lie groups and their non-compact duals},
Differential Geom. Appl. {\bf 24} (2006), 351-366.

[26]
S. Gudmundsson, M. Svensson,
{\it On the existence of harmonic morphisms from certain symmetric spaces},
J. Geom. Phys. {\bf 57} (2007), 353-366.

[27]
S. Gudmundsson, M. Svensson,
{\it Harmonic morphisms from solvable Lie groups},
Math. Proc. Cambridge Philos. Soc. {\bf 147} (2009), 389-408.

[28]
S. Gudmundsson, M. Svensson,
{\it Harmonic morphisms from homogeneous spaces of positive curvature},
Math. Proc. Cambridge Philos. Soc. {\bf 157} (2014), 321-327.

[29]
S. Gudmundsson, M. Svensson, M. Ville,
{\it Complete minimal submanifolds of compact Lie groups},
Math. Z. {\bf 282} (2016), 993-1005.

[30]
H. Haber,
{\it Notes on Antisymmetric Matrices and the Pfaffian}, Lecture Notes, University of Califoria, Santa Cruz (2015),\\
\href{https://scipp.ucsc.edu/~haber/webpage/pfaffian2.pdf}{\tt www.scipp.ucsc.edu/~haber/webpage/pfaffian2.pdf}

[31]
B.C. Hall,
{\it Lie Groups, Lie Algebras, and Representations}, Second Edition, Graduate Texts in Mathematics {\bf 222}, Springer (2015).

[32]
S. Helgason,
{\it Differential Geometry, Lie Groups, and Symmetric Spaces},
Graduate Studies in Mathematics {\bf 34}
American Mathematical Society (1978).

[33]
W. Y. Hsiang, H. B. Lawson,
{\it Minimal submanifolds of low cohomogeneity},
J. Diff. Geom. {\bf 5} (1971), 1–38.

[34]
A.W. Knapp,
{\it Lie Groups Beyond an Introduction}, Progress in Mathematics {\bf 140}, Birkhäuser (2002).

[35]
S. Kobayashi, K.Nomizu,
{\it Foundations of Differential Geometry, Volume 1},
Interscience Tracts in Pure and Applied Mathematics {\bf 15},
Interscience Publishers (1963).

[36]
J. L. Lagrange, {\it Essai d’une nouvelle méthode pour determiner les maxima et les minima des formules integrales indefinies}, Miscellanea Taurinensia 2 325 (1760).

[37]
H.B. Lawson,
{\it Complete minimal surfaces in $S^3$}, Ann. of Math. {\bf 92} (1970), 335–74.

[38]
J.M. Lee,
{\it Introduction to Smooth Manifolds},
Graduate Texts in Mathematics {\bf 218}
Springer (2013).

[39]
A. Lindström,
{\it Harmonic Morphisms and p-Harmonic Functions on the Classical Compact Symmetric Spaces via the Cartan Embedding}, Master's Thesis, Lund University (2023),\\
\href{https://arxiv.org/abs/2402.15532}{\tt www.arxiv.org/abs/2402.15532}

[40]
J. Milnor, {\it Singular Points of Complex Hypersurfaces}, Annals of Mathematical Studies {\bf 61} (1968).

[41]
S.B. Myers, N.E. Steenrod,
{ \it The group of isometries of a Riemannian manifold},
Ann. Math., {\bf 40} (1939), 400-416.

[42]
B. O'Neill,
{\it Semi-Riemannian Geometry with Applications to Relativity}, Academic Press (1983).

[43]
O. Riedler, A. Siffert,
{\it Global eigenfamilies on compact manifolds}, preprint, (2024).

[44]
M. Svensson,
{\it Harmonic Morphisms, Hermitian Structures and Symmetric Spaces}, Doctoral Thesis, Lund University (2004),\\
\href{https://www.matematik.lu.se/matematiklu/personal/sigma/students/Martin-Svensson-PhD.pdf}{\tt www.matematik.lu.se/matematiklu/personal/sigma/students/Martin-Svensson-PhD.pdf}

\end{thebibliography}
\backcover

\end{document}