$\displaystyle\mathbb{E}|A_{\gamma}|=\mathbb{E}|U_{\gamma}|$
$\displaystyle\leq\mathbb{E}\sum_{k=-\infty}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{1}{3}k/m}\leq\tau_{1}.$

Part 2. Note that
$\sum_{k=-\infty}^{a-1}e^{\frac{1}{3}k/m}=e^{\frac{1}{3}(a-1)/m}\frac{1}{1-e^{-1/(3m)}}.$
By Lemma C.1(2),

$\displaystyle\left|U_{\gamma;-\infty,a}+e^{\frac{1}{3}(a-1)/m}\frac{1}{1-e^{-1/(3m)}}\right|$
$\displaystyle=\left|\sum_{k=-\infty}^{a-1}({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)e^{\frac{1}{3}k/m}+\sum_{k=-\infty}^{a-1}e^{\frac{1}{3}k/m}\right|$
$\displaystyle=\left|\sum_{k=-\infty}^{a-1}{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}e^{\frac{1}{3}k/m}\right|\leq\tau_{2}.$

Part 3. By Lemma C.1(3),

$\displaystyle\mathbb{E}|U_{\gamma;b,\infty}|$
$\displaystyle\leq\mathbb{E}\sum_{k=b}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{1}{3}k/m}\leq\tau_{3}.$

Part 4. Follows from parts 1 and 2.

Part 5. Follows from parts 2 and 3.

Part 6. Follows from part 2.

Part 7. Recall that $X_{j},X_{k}$ are independent for $j\neq k$.

$\displaystyle\mathbb{E}|C_{\gamma}||C_{\gamma^{\prime}}|=\mathbb{E}|U_{\gamma;b,\infty}U_{\gamma^{\prime};b,\infty}|$
$\displaystyle=\mathbb{E}\left|\sum_{k=b}^{\infty}\sum_{j=b}^{\infty}({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{j}},{\gamma^{\prime}}\boldsymbol{)}}-1)e^{\frac{k+j}{3m}}\right|$
$\displaystyle\leq\sum_{k=b}^{\infty}\sum_{j=b}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{j}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{k+j}{3m}}$
$\displaystyle\leq\tau_{3}^{2}+\sum_{k=b}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{k}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{2k}{3m}},$

where by Lemma C.1(3) we know $\tau_{3}$ upper bounds
$\mathbb{E}\sum_{k=b}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{1}{3}k/m}$.
Note that $|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|\leq 2$ and
$|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|=0$ if $X_{k}=0$. We also
have

$\displaystyle\sum_{k=b}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{k}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{2k}{3m}}$
$\displaystyle\leq 4\sum_{k=b}^{\infty}\mathbbm{1}\left[{X_{k}\neq
0}\right]e^{\frac{2k}{3m}}$ $\displaystyle=4\sum_{k=b}^{\infty}(1-e^{-\lambda
e^{-k/m}})e^{\frac{2}{3}k/m}$ $\displaystyle\leq 4\sum_{k=b}^{\infty}\lambda
e^{-k/m}\cdot e^{\frac{2}{3}k/m}$ $\displaystyle=4\lambda
e^{-\frac{1}{3}b/m}\frac{1}{1-e^{-1/(3m)}}\;\leq\;16m\lambda
e^{-\frac{1}{3}b/m}.$

The last inequality follows from $e^{-1/(3m)}\leq 1-\frac{1}{4m}$ which holds
when $m\geq 2$. We conclude that

$\displaystyle\mathbb{E}|U_{\gamma;b,\infty}U_{\gamma^{\prime};b,\infty}|$
$\displaystyle\leq\tau_{3}^{2}+16m\lambda e^{-\frac{1}{3}b/m}.$

Part 8. Applying parts 1 and 3, we have

$\displaystyle\mathbb{E}|A\gamma||C_{\gamma^{\prime}}|=\mathbb{E}|U_{\gamma}U_{\gamma^{\prime};b,\infty}|$
$\displaystyle=\mathbb{E}\left|\sum_{k=-\infty}^{\infty}\sum_{j=b}^{\infty}({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{j}},{\gamma^{\prime}}\boldsymbol{)}}-1)e^{\frac{k+j}{3m}}\right|$
$\displaystyle\leq\sum_{k=-\infty}^{\infty}\sum_{j=b}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{j}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{k+j}{3m}}$
$\displaystyle\leq\tau_{1}\tau_{3}+\sum_{k=b}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{k}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{2k}{3m}},$
where by Lemma C.1 we know $\tau_{1}$ upper bounds
$\mathbb{E}\sum_{k=-\infty}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{1}{3}k/m}$
and $\tau_{3}$ upper bounds
$\mathbb{E}\sum_{k=b}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{1}{3}k/m}$.
From the proof of part 7, we know this is
$\displaystyle\leq\tau_{1}\tau_{3}+16m\lambda e^{-\frac{1}{3}b/m}.$

Part 9. Applying part 1,

$\displaystyle\mathbb{E}|A_{\gamma}||A_{\gamma^{\prime}}|=\mathbb{E}|U_{\gamma}U_{\gamma^{\prime}}|$
$\displaystyle=\mathbb{E}\left|\sum_{k=-\infty}^{\infty}\sum_{j=-\infty}^{\infty}({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{j}},{\gamma^{\prime}}\boldsymbol{)}}-1)e^{\frac{k+j}{3m}}\right|$
$\displaystyle\leq\sum_{k=-\infty}^{\infty}\sum_{j=-\infty}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{j}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{k+j}{3m}}$
$\displaystyle\leq\tau_{1}^{2}+\sum_{k=-\infty}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1)({\boldsymbol{(}{X_{k}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{2k}{3m}}$
where by Lemma C.1 we know $\tau_{1}$ upper bounds
$\mathbb{E}\sum_{k=-\infty}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{1}{3}k/m}$.
$\displaystyle\leq\tau_{1}^{2}+2\sum_{k=-\infty}^{\infty}\mathbb{E}\left|({\boldsymbol{(}{X_{k}},{\gamma^{\prime}}\boldsymbol{)}}-1)\right|e^{\frac{2k}{3m}}.$

Similar to the proof of Lemma C.1(1), we have

$\displaystyle\mathbb{E}\sum_{k=-\infty}^{\infty}|{\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}}-1|e^{\frac{2}{3}k/m}$
$\displaystyle\leq
2m(a^{c}-b^{c})\Gamma(-c)+2\left(2\left(1+\frac{c}{1-c}\right)|b-a|^{c}+(|a|^{c}+|b|^{c})\Gamma(1-c)\right)$
where $a=0,b=\lambda,c=2/3$
$\displaystyle=2m\lambda^{2/3}(-\Gamma(-2/3))+2\lambda^{2/3}(6+\Gamma(1/3))\;\leq\;18m\lambda^{2/3}.$

Where the last inequality holds when $m\geq 2$. We conclude that
$\mathbb{E}|U_{\gamma}U_{\gamma^{\prime}}|\leq\tau_{1}^{2}+36m\lambda^{2/3}$.
∎

We are now in a position to prove Theorem 6.1.

###### Proof of Theorem 6.1.

Part 1: bias. Note that $U_{\gamma}=A_{\gamma}$ and
$U_{\gamma;a,b}-\tau_{2}=A_{\gamma}-B_{\gamma}-C_{\gamma}$.

$\displaystyle|\mathbb{E}\phi_{f,m}(X)-\mathbb{E}\phi_{f,m;a,b}(X)|$
$\displaystyle\leq\mathbb{E}|\phi_{f,m}(X)-\phi_{f,m;a,b}(X)|$
$\displaystyle=\mathbb{E}\left|\int{\Gamma}\hat{f}(\gamma)\frac{\prod_{j=1}^{3}A_{\gamma}^{(j)}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})}{m^{3}\Gamma(-1/3)^{3}}\,d\mu_{\mathbbold{\Gamma}}(\gamma)\right|$
$\displaystyle\leq\int{\Gamma}|\hat{f}(\gamma)|\frac{\mathbb{E}\left|\prod_{j=1}^{3}A_{\gamma}^{(j)}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\right|}{m^{3}|\Gamma(-1/3)|^{3}}\,d\mu{\Gamma}\gamma.$

By the independence of the three copies,

$\displaystyle\mathbb{E}\left|\prod_{j=1}^{3}A_{\gamma}^{(j)}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\right|$
$\displaystyle\leq
3(\mathbb{E}|A_{\gamma}|)^{2}(\mathbb{E}|B_{\gamma}|+\mathbb{E}|C_{\gamma}|)+3(\mathbb{E}|A_{\gamma}|)(\mathbb{E}|B_{\gamma}|+\mathbb{E}|C_{\gamma}|)^{2}+(\mathbb{E}|B_{\gamma}|+\mathbb{E}|C_{\gamma}|)^{3}.$

Since we assumed $a\leq m(\ln\lambda-6\ln m)$ and $b\geq m(\ln\lambda+3\ln
m)$, by Lemma C.2 we have

$\displaystyle\mathbb{E}|A_{\gamma}|$
$\displaystyle\leq\tau_{1}=16m\lambda^{1/3},$
$\displaystyle\mathbb{E}|B_{\gamma}|$ $\displaystyle\leq\tau_{2}\leq
4me^{\frac{1}{3}a/m}\leq 4m^{-1}\lambda^{1/3},$
$\displaystyle\mathbb{E}|C_{\gamma}|$ $\displaystyle\leq\tau_{3}=4m\lambda
e^{-\frac{2}{3}b/m}\leq 4m^{-1}\lambda^{1/3}.$

Thus we have

$\displaystyle\mathbb{E}\left|\prod_{j=1}^{3}A_{\gamma}^{(j)}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\right|$
$\displaystyle=O(m^{2}\lambda^{2/3}\cdot m^{-1}\lambda^{1/3})=\lambda\cdot
O(m).$

We conclude that

$\displaystyle|\mathbb{E}\phi_{f,m}(X)-\mathbb{E}\phi_{f,m;a,b}(X)|$
$\displaystyle\leq\int{\Gamma}|\hat{f}(\gamma)|\frac{\lambda\cdot
O(m)}{m^{3}|\Gamma(-1/3)|^{3}}\,d\mu{\Gamma}\gamma$
$\displaystyle=\lambda\|\hat{f}\|_{1}O(m^{-2}).$

Part 2: variance.

$\displaystyle\mathbb{E}\left|\phi_{f,m}(X)\overline{\phi_{f,m}(X)}-\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}\right|$
$\displaystyle\leq\mathbb{E}\left|\int{\Gamma}\int{\Gamma}\hat{f}(\gamma)\overline{\hat{f}(\gamma)}\frac{\prod_{j=1}^{3}A_{\gamma}^{(j)}\overline{A_{\gamma^{\prime}}^{(j)}}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\overline{(A_{\gamma^{\prime}}^{(j)}-B_{\gamma^{\prime}}^{(j)}-C_{\gamma^{\prime}}^{(j)})}}{m^{6}\Gamma(-1/3)^{6}}\,d\mu_{\mathbbold{\Gamma}}(\gamma)d\mu_{\mathbbold{\Gamma}}(\gamma^{\prime})\right|$
$\displaystyle\leq\int{\Gamma}\int{\Gamma}|\hat{f}(\gamma)||\overline{\hat{f}(\gamma)}|\frac{\mathbb{E}\left|\prod_{j=1}^{3}A_{\gamma}^{(j)}\overline{A_{\gamma^{\prime}}^{(j)}}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\overline{(A_{\gamma^{\prime}}^{(j)}-B_{\gamma^{\prime}}^{(j)}-C_{\gamma^{\prime}}^{(j)})}\right|}{m^{6}|\Gamma(-1/3)|^{6}}\,d\mu_{\mathbbold{\Gamma}}(\gamma)d\mu_{\mathbbold{\Gamma}}(\gamma^{\prime}),$

We expand the product
$\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\overline{(A_{\gamma^{\prime}}^{(j)}-B_{\gamma^{\prime}}^{(j)}-C_{\gamma^{\prime}}^{(j)})}$
and cancel the term
$\prod_{j=1}^{3}A_{\gamma}^{(j)}\overline{A_{\gamma^{\prime}}^{(j)}}$. By the
independence of the three copies, we have

$\displaystyle\mathbb{E}\left|\prod_{j=1}^{3}A_{\gamma}^{(j)}\overline{A_{\gamma^{\prime}}^{(j)}}-\prod_{j=1}^{3}(A_{\gamma}^{(j)}-B_{\gamma}^{(j)}-C_{\gamma}^{(j)})\overline{(A_{\gamma^{\prime}}^{(j)}-B_{\gamma^{\prime}}^{(j)}-C_{\gamma^{\prime}}^{(j)})}\right|$
$\displaystyle\leq(z_{AA}+z_{BB}+z_{CC}+2z_{AB}+2z_{BC}+2z_{CA})^{3}-z_{AA}^{3}$

where $z_{AA},z_{BB},z_{CC},z_{AB},z_{BC},z_{CA}$ are the upper bounds from
Lemma C.2. Since we assumed $a\leq m(\ln\lambda-6\ln m)$ and $b\geq
m(\ln\lambda+3\ln m)$, we have

$\displaystyle\mathbb{E}|A_{\gamma}||A_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{AA}=\tau_{1}^{2}+36m\lambda^{2/3}=\lambda^{2/3}O(m^{2}),$
$\displaystyle\mathbb{E}|B_{\gamma}||B_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{BB}=\tau_{2}^{2}=\lambda^{2/3}O(m^{-2}),$
$\displaystyle\mathbb{E}|C_{\gamma}||C_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{CC}=\tau_{3}^{2}+16m\lambda e^{-\frac{1}{3}b/m}=\lambda^{2/3}O(1),$
$\displaystyle\mathbb{E}|A_{\gamma}||B_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{AB}=\tau_{1}\tau_{2}=\lambda^{2/3}O(1),$
$\displaystyle\mathbb{E}|B_{\gamma}||C_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{BC}=\tau_{2}\tau_{3}=\lambda^{2/3}O(m^{-2}),$
$\displaystyle\mathbb{E}|C_{\gamma}||A_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{CA}=\tau_{1}\tau_{3}+16m\lambda e^{-\frac{1}{3}b/m}=\lambda^{2/3}O(1).$

Thus we have

$\displaystyle(z_{AA}+z_{BB}+z_{CC}+2z_{AB}+2z_{BC}+2z_{CA})^{3}-z_{AA}^{3}=\lambda^{2}O(m^{4}),$

and therefore

$\displaystyle\mathbb{E}\left|\phi_{f,m}(X)\overline{\phi_{f,m}(X)}-\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}\right|$
$\displaystyle\leq\lambda^{2}\|\hat{f}\|_{1}^{2}O(m^{-2}).$

We conclude that

$\displaystyle|\mathbb{V}(\phi_{f,m;a,b}(X))-\mathbb{V}(\phi_{f,m;}(X))|$
$\displaystyle\leq\left|\mathbb{E}\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}-(\mathbb{E}\phi_{f,m;a,b}(X))\overline{(\mathbb{E}\phi_{f,m;a,b}(X))}\right.$
$\displaystyle\quad\quad\left.{}-\mathbb{E}\phi_{f,m}(X)\overline{\phi_{f,m}(X)}+(\mathbb{E}\phi_{f,m}(X))\overline{(\mathbb{E}\phi_{f,m}(X))}\right|$
Note that for two complex numbers $w_{1},w_{2}$,
$|w_{1}\overline{w_{1}}-w_{2}\overline{w_{2}}|\leq|w_{1}-w_{2}|(|w_{1}|+|w_{2}|)$
$\displaystyle\leq\mathbb{E}\left|\phi_{f,m}(X)\overline{\phi_{f,m}(X)}-\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}\right|$
$\displaystyle\quad\quad{}+\mathbb{E}|\phi_{f,m}(X)-\phi_{f,m;a,b}(X)|(|\mathbb{E}\phi_{f,m}(X)|+|\mathbb{E}\phi_{f,m;a,b}(X)|)$
$\displaystyle=\lambda^{2}\|\hat{f}\|_{1}^{2}O(m^{-2})+\lambda\|\hat{f}\|_{1}O(m^{-2})(\lambda|\mathbb{E}f(R)|+\lambda\|\hat{f}\|_{1}O(m^{-1}))$
where $|\mathbb{E}f(R)|\leq\|f\|_{\infty}$ and, by Lemma 2.4,
$\|f\|_{\infty}\leq\|\hat{f}\|_{1}$. In conclusion, this is
$\displaystyle=\lambda^{2}\|\hat{f}\|_{1}^{2}O(m^{-2}).\hskip 204.85974pt\qed$

##  Appendix D Proof of Theorem 6.2 — Depoissonization

Let use recall Theorem 6.2.

See 6.2

In the analysis of truncated estimator, we implicitly used a natural coupling
between the Poisson tower $(X_{k})_{k=-\infty}^{\infty}$ and the truncated
Poisson tower $(X_{k})_{k=a}^{b-1}$, where both share the same outcomes in the
overlapping indices. For the analysis of depoissonization, we need to
construct an explicit coupling $(X_{k},\tilde{X}_{k})_{k=a}^{b-1}$ where
$X_{k}$s are from the Poisson tower while $\tilde{X}_{k}$s are from the
binomial tower.

######  Lemma D.1 (coupling).

Fix a state vector $\mathbf{x}$. There exists a random process
$(W_{\gamma},Z_{\gamma},K_{k},M_{k,j})_{\gamma\in\mathbbold{\Gamma},k\in[\lambda],j\in\mathbb{Z}_{+}}$
such that

$\displaystyle(W_{\gamma})_{\gamma\in\mathbbold{\Gamma}}$
$\displaystyle\sim\left(\sum_{k=a}^{b-1}(\boldsymbol{(}{X_{k}},{\gamma}\boldsymbol{)}-1)e^{\frac{1}{3}k/m}-e^{\frac{1}{3}(a-1)/m}\frac{1}{1-e^{-1/(3m)}}\right)_{\gamma\in\mathbbold{\Gamma}}$
$\displaystyle(W_{\gamma}+Z_{\gamma})_{\gamma\in\mathbbold{\Gamma}}$
$\displaystyle\sim\left(\sum_{k=a}^{b-1}(\boldsymbol{(}{\tilde{X_{k}}},{\gamma}\boldsymbol{)}-1)e^{\frac{1}{3}k/m}-e^{\frac{1}{3}(a-1)/m}\frac{1}{1-e^{-1/(3m)}}\right)_{\gamma\in\mathbbold{\Gamma}}$

and for any $\gamma$

$\displaystyle|Z_{\gamma}|$ $\displaystyle\leq
2\sigma_{2}+2\sum_{k=1}^{\lambda}\left(\mathbbm{1}\left[{K_{k}\geq
2}\right]\sum_{j=1}^{K_{k}-1}e^{\frac{1}{3}M_{j,k}/m}+\mathbbm{1}\left[{K_{k}=-1}\right]e^{\frac{1}{3}M_{1,k}/m}\right)$
$\displaystyle|W_{\gamma}|$ $\displaystyle\leq
2\sigma_{2}+2\sum_{k=1}^{\lambda}\sum_{j=1}^{K_{k}}e^{\frac{1}{3}M_{j,k}/m}$

where the $\\{K_{k}\\}$ are i.i.d. copies of $K$ and the $\\{M_{j,k}\\}$ are
i.i.d copies of $M$, defined as follows.

$\displaystyle\mathbb{P}(K=-1)$ $\displaystyle=e^{-\sigma_{1}}-(1-\sigma_{1})$
$\displaystyle\mathbb{P}(K=0)$ $\displaystyle=1-\sigma_{1}$
$\displaystyle\mathbb{P}(K=k)$
$\displaystyle=e^{-\sigma_{1}}\frac{\sigma_{1}^{k}}{k!},\quad\forall k\geq 1.$
$\displaystyle\mathbb{P}(M=r)$
$\displaystyle=\frac{e^{-r/m}}{\sigma_{1}},\quad\forall
r\in\\{t,\ldots,b-1\\},$

where $t=m(\ln\lambda-6\ln m)$ and

$\displaystyle\sigma_{1}$
$\displaystyle=\sum_{k=t}^{\infty}e^{-k/m}=O(m)e^{-t/m}=\lambda^{-1}O(m^{7}),$
$\displaystyle\sigma_{2}$
$\displaystyle=\sum_{k=\infty}^{t-1}e^{\frac{1}{3}k/m}\leq
O(m)e^{\frac{1}{3}t/m}=\lambda^{1/3}O(m^{-1}).$

###### Proof.

Figure 9: Coupling between Bernoulli and Poisson random variables with the
same mean.

Without loss of generality, assume
$\mathbf{x}=\\{x_{1},x_{2},\ldots,x_{\lambda},0,0,\ldots\\}$ where
$x_{1},\ldots,x_{\lambda}$ are the only non-zeros and $\lambda$ is the support
size.

Initialize two vectors $(y_{k})_{k=a,\ldots,b-1}$ and
$(z_{k})_{k=a,\ldots,b-1}$ with zeros. We are going to update the two vectors
through the following process and the final state of
$(y_{k})_{k=a,\ldots,b-1}$ will distribute as
$(\tilde{X}_{k})_{k=a,\ldots,b-1}$ and the final state of
$(z_{k})_{k=a,\ldots,b-1}$ will distribute as $(X_{k})_{k=a,\ldots,b-1}$.

#### Generation of shared randomness.

Generate the variables $K_{k}$ and $M_{j,k}$ for all $k\in[\lambda]$ and
$j\in\mathbb{Z}_{+}$, where $K_{k}$s are i.i.d. copies of $K$ and $M_{j,k}$s
are i.i.d. copies of $M$. The distribution of $K$ and $M$ are defined in the
theorem statement.

#### Generation of $(y_{k})_{k=a,\ldots,b-1}$.

  * •

For $k=1,\ldots,\lambda$,

    * –

If $K_{k}=0$,

      * *

Randomly sample $L_{k}$ over $\\{a,\ldots,t-1\\}$ where
$\mathbb{P}(L_{k}=w)\propto e^{-w/m}$.

      * *

$y_{L_{k}}\leftarrow y_{L_{k}}+x_{k}$

    * –

If $K_{k}\neq 0$,

      * *

$y_{M_{1,k}}\leftarrow y_{M_{1,k}}+x_{k}$

#### Generation of $(z_{k})_{k=a,\ldots,b-1}$.

  * •

Randomly sample $(z_{k})_{k=a,\ldots,t-1}$ according to the distribution of
$(X_{k})_{k=a,\ldots,t-1}$.

  * •

For $k=1,\ldots,\lambda$,

    * –

If $K_{k}\geq 1$, for $j=1,\ldots,K_{k}$,

      * *

$z_{M_{j,k}}\leftarrow z_{M_{j,k}}+x_{k}$.

It is clear by construction that at the end of the procedures,
$(y_{k})_{k=a,\ldots,b-1}$ distributes as $(\tilde{X}_{k})_{k=a,\ldots,b-1}$
and $(z_{k})_{k=a,\ldots,b-1}$ distributes as $(X_{k})_{k=a,\ldots,b-1}$. In
the same time, they are coupled in the way their cells between $t$ to $b-1$
look almost the same.

Now for any $\gamma\in\mathbbold{\Gamma}$,

$\displaystyle W_{\gamma}$
$\displaystyle=\sum_{k=a}^{b-1}(\boldsymbol{(}{z_{k}},{\gamma}\boldsymbol{)}-1)e^{\frac{1}{3}k/m}-e^{\frac{1}{3}(a-1)/m}\frac{1}{1-e^{-1/(3m)}},$
$\displaystyle Z_{\gamma}$
$\displaystyle=\sum_{k=a}^{b-1}(\boldsymbol{(}{y_{k}},{\gamma}\boldsymbol{)}-1)e^{\frac{1}{3}k/m}-e^{\frac{1}{3}(a-1)/m}\frac{1}{1-e^{-1/(3m)}}-W_{\gamma}$
$\displaystyle=\sum_{k=a}^{b-1}(\boldsymbol{(}{y_{k}},{\gamma}\boldsymbol{)}-\boldsymbol{(}{z_{k}},{\gamma}\boldsymbol{)})e^{\frac{1}{3}k/m}.$

Then

$\displaystyle\mathbb{E}|W_{\gamma}|$ $\displaystyle\leq
2\sum_{k=-\infty}^{t-1}e^{\frac{1}{3}k/m}+2\sum_{k=t}^{b-1}\mathbbm{1}\left[{z_{k}\neq
0}\right]e^{\frac{1}{3}k/m}.$ A cell is non-zero only if it is hit by at least
one element, thus $\displaystyle\leq
2\sum_{k=-\infty}^{t-1}e^{\frac{1}{3}k/m}+2\sum_{k=1}^{\lambda}\sum_{j=1}^{K_{k}}e^{\frac{1}{3}M_{j,k}/m}.$

The difference $Z_{\lambda}$ can be bound similarly. First, for all cells from
$a$ to $t-1$, we bound the difference
$\sum_{k=a}^{t-1}|\boldsymbol{(}{y_{k}},{\gamma}\boldsymbol{)}-\boldsymbol{(}{z_{k}},{\gamma}\boldsymbol{)}|e^{\frac{1}{3}k/m}$
by $2\sigma_{2}$.

Then we focus on cells from $t$ to $b-1$. We bound the difference
$\sum_{k=t}^{b-1}|\boldsymbol{(}{y_{k}},{\gamma}\boldsymbol{)}-\boldsymbol{(}{z_{k}},{\gamma}\boldsymbol{)}|e^{\frac{1}{3}k/m}$
by $2\sum_{j=t}^{b-1}\mathbbm{1}\left[{y_{j}\neq
z_{j}}\right]e^{\frac{1}{3}j/m}$. Now we track
$\sum_{j=t}^{b-1}\mathbbm{1}\left[{y_{j}\neq z_{j}}\right]e^{\frac{1}{3}j/m}$,
step by step. Initially, they are the same since they are all zeros. For each
element $k$,

  * •

if $K_{k}=0$, the cells from $t$ to $b-1$ won’t be changed in both $y$ and
$z$;

  * •

if $K_{k}=1$, exactly one cell, the cell $M_{1,k}$, is incremented by $x_{k}$
in both $y$ and $z$, introducing no difference;

  * •

if $K_{k}=-1$, the cell $M_{1,k}$, is incremented by $x_{k}$ in both $y$ but
not in $z$, introducing at most a difference of size
$e^{\frac{1}{3}M_{1,k}/3}$;

  * •

if $K_{k}\geq 2$, the cells $M_{j,k}$, for $j=2,\ldots,K_{k}$ are incremented
by $x_{k}$ in $z$ but not in $y$, introducing at most a difference of size
$\sum_{j=2}^{K_{k}}e^{\frac{1}{3}M_{j,k}/3}$.

Thus,

$\displaystyle\mathbb{E}|Z_{\gamma}|$ $\displaystyle\leq
2\sigma_{2}+2\sum_{k=1}^{\lambda}\left(\mathbbm{1}\left[{K_{k}\geq
2}\right]\sum_{j=1}^{K_{k}-1}e^{\frac{1}{3}M_{j,k}/m}+\mathbbm{1}\left[{K_{k}=-1}\right]e^{\frac{1}{3}M_{1,k}/m}\right).$

∎

######  Lemma D.2.

The following statements are true. If $\lambda=\Omega(m^{13})$, then for any
$\gamma,\gamma^{\prime}\in\mathbbold{\Gamma}$,

  1. 1.

$\mathbb{E}|W_{\gamma}|=\lambda^{1/3}O(m)$.

  2. 2.

$\mathbb{E}|W_{\gamma}||W_{\gamma^{\prime}}|=\lambda^{2/3}O(m^{2})$.

  3. 3.

$\mathbb{E}|Z_{\gamma}|=\lambda^{1/3}O(m^{-1})$.

  4. 4.

$\mathbb{E}|W_{\gamma}||Z_{\gamma^{\prime}}|=\lambda^{2/3}O(1)$.

  5. 5.

$\mathbb{E}|Z_{\gamma}||Z_{\gamma^{\prime}}|=\lambda^{2/3}O(m^{-2})$.

  6.

###### Proof.

Since $\lambda=\Omega(m^{13})$, we can assume
$\sigma_{1}=\lambda^{-1}O(m^{7})\leq 1$. Recall $t=m(\ln\lambda-6\ln m)$.
Compute

$\displaystyle\mathbb{E}e^{\frac{1}{3}M/m}$
$\displaystyle=\sum_{r=t}^{b-1}\frac{e^{-r/m}}{\sigma_{1}}e^{\frac{1}{3}r/m}=O(1)\frac{e^{-\frac{2}{3}t/m}-e^{-\frac{2}{3}b/m}}{e^{-t/m}-e^{-b/m}}=\lambda^{1/3}O(m^{-2})$
(9) $\displaystyle\mathbb{E}e^{\frac{2}{3}M/m}$
$\displaystyle=\sum_{r=t}^{b-1}\frac{e^{-r/m}}{\sigma_{1}}e^{\frac{2}{3}r/m}=O(1)\frac{e^{-\frac{1}{3}t/m}-e^{-\frac{1}{3}b/m}}{e^{-t/m}-e^{-b/m}}=\lambda^{2/3}O(m^{-4})\
$ (10)

Part 1. Follows from Lemma C.2.

Part 2. Follows from Lemma C.2.

Part 3.

$\displaystyle\mathbb{E}|Z_{\gamma}|$ $\displaystyle\leq
2\sigma_{2}+2\mathbb{E}\sum_{k=1}^{\lambda}\left(\mathbbm{1}\left[{K_{k}\geq
2}\right]\sum_{j=1}^{K_{k}-1}e^{\frac{1}{3}M_{j,k}/m}+\mathbbm{1}\left[{K_{k}=-1}\right]e^{\frac{1}{3}M_{1,k}/m}\right)$
$\displaystyle=\lambda^{1/3}O(m^{-1})+2\lambda\sum_{k=2}^{\infty}\mathbb{P}(K=k)(k-1)\mathbb{E}e^{\frac{1}{3}M/m}+2\lambda\mathbb{P}(K=-1)\mathbb{E}e^{\frac{1}{3}M/m},$
where $\sum_{k=2}^{\infty}\mathbb{P}(K=k)(k-1)=\mathbb{E}(K-1;K\geq
2)=\sigma_{1}-1+e^{-\sigma_{1}}=O(\sigma_{1}^{2})$ and
$\mathbb{P}(K=-1)=\sigma_{1}-1+e^{-\sigma_{1}}=O(\sigma_{1}^{2})$. By Eq. 9,
we know $\mathbb{E}e^{\frac{1}{3}M/m}=\lambda^{1/3}O(m^{-2}).$
$\displaystyle=\lambda^{1/3}O(m^{-1})+\lambda\cdot\lambda^{1/3}O(m^{-2})O(\sigma_{1}^{2})$
recall $\sigma_{1}=\lambda^{-1}(m^{7})$
$\displaystyle=\lambda^{1/3}O(m^{-1})+\lambda^{1/3}\frac{O(m^{12})}{\lambda}$
which, since we assumed $\lambda=\Omega(m^{13})$, is
$\displaystyle=\lambda^{1/3}O(m^{-1}).$

Part 4.

$\displaystyle\mathbb{E}|W_{\gamma}||Z_{\gamma^{\prime}}|$
$\displaystyle\leq\mathbb{E}|W_{\gamma}|\mathbb{E}|Z_{\gamma^{\prime}}|+4\mathbb{E}\sum_{k=1}^{\lambda}\left(\mathbbm{1}\left[{K_{k}\geq
2}\right]\sum_{j=1}^{K_{k}-1}e^{\frac{1}{3}M_{j,k}/m}+\mathbbm{1}\left[{K_{k}=-1}\right]e^{\frac{1}{3}M_{1,k}/m}\right)\sum_{j=1}^{K_{k}}e^{\frac{1}{3}M_{j,k}/m}$
$\displaystyle=\lambda^{2/3}O(1)+4\lambda\mathbb{E}\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j,k}/m}\sum_{j=1}^{K}e^{\frac{1}{3}M_{j,k}/m}$
$\displaystyle=\lambda^{2/3}O(1)+4\lambda\mathbb{E}(\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j}/m})^{2}+4\lambda\mathbb{E}(K-1;K\geq
2)(\mathbb{E}e^{\frac{1}{3}M/m})^{2}.$

We estimate

$\displaystyle\mathbb{E}(\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j}/m})^{2}$
$\displaystyle\leq\left(\mathbb{E}\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j,k}/m}\right)^{2}+\mathbb{E}\sum_{j=1}^{K-1}e^{\frac{2}{3}M_{j,k}/m}$
$\displaystyle=(\mathbb{E}(K-1;K\geq
2)(\mathbb{E}e^{\frac{1}{3}M/m}))^{2}+\mathbb{E}(K-1;K\geq
2)(\mathbb{E}e^{\frac{2}{3}M/m})$ We have shown that $\mathbb{E}(K-1;K\geq
2)=O(\sigma_{1}^{2})$, $\mathbb{E}e^{\frac{1}{3}M/m}=\lambda^{1/3}O(m^{-2})$
and $\mathbb{E}e^{\frac{2}{3}M/m}=\lambda^{2/3}O(m^{-4})$.
$\displaystyle=\sigma_{1}^{4}\lambda^{2/3}O(m^{-4})+\sigma_{1}^{2}\lambda^{2/3}O(m^{-4})$
$\displaystyle=\sigma_{1}^{2}\lambda^{2/3}O(m^{-4}),$

since $\sigma_{1}<1$. Thus,

$\displaystyle\mathbb{E}|W_{\gamma}||Z_{\gamma^{\prime}}|$
$\displaystyle=\lambda^{2/3}O(1)+\lambda\sigma_{1}^{2}\lambda^{2/3}O(m^{-4})+\lambda\sigma_{1}^{2}\lambda^{2/3}O(m^{-4})$
$\displaystyle=\lambda^{2/3}O(1)+\lambda^{2/3}\frac{O(m^{-10})}{\lambda}$ and
since we assumed $\lambda=\Omega(m^{13})$, $\displaystyle=\lambda^{2/3}O(1).$

Part 5.

$\displaystyle\mathbb{E}|Z_{\gamma}||Z_{\gamma^{\prime}}|$
$\displaystyle\leq\mathbb{E}|Z_{\gamma}|\mathbb{E}|Z_{\gamma^{\prime}}|+4\mathbb{E}\sum_{k=1}^{\lambda}\left(\mathbbm{1}\left[{K_{k}\geq
2}\right]\sum_{j=1}^{K_{k}-1}e^{\frac{1}{3}M_{j,k}/m}+\mathbbm{1}\left[{K_{k}=-1}\right]e^{\frac{1}{3}M_{1,k}/m}\right)^{2}$
$\displaystyle=\lambda^{2/3}O(m^{-2})+4\lambda\mathbb{E}\left(\mathbbm{1}\left[{K\geq
2}\right]\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j}/m}+\mathbbm{1}\left[{K=-1}\right]e^{\frac{1}{3}M_{1}/m}\right)^{2}$
$\displaystyle=\lambda^{2/3}O(m^{-2})+4\lambda\mathbb{P}(K=-1)\mathbb{E}e^{\frac{2}{3}M/m}+4\lambda\mathbb{E}(\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j}/m})^{2}$
We have shown that $\mathbb{P}(K=-1)=O(\sigma_{1}^{2})$,
$\mathbb{E}e^{\frac{1}{3}M/m}=\lambda^{1/3}O(m^{-2})$ and
$\mathbb{E}(\sum_{j=1}^{K-1}e^{\frac{1}{3}M_{j}/m})^{2}=\sigma_{1}^{2}\lambda^{2/3}O(m^{-4})$.
$\displaystyle=\lambda^{2/3}O(m^{-2})+\lambda\sigma_{1}^{2}\lambda^{2/3}O(m^{-4})+\lambda\sigma_{1}^{2}\lambda^{2/3}O(m^{-4})$
$\displaystyle=\lambda^{2/3}O(m^{-2})+\lambda^{2/3}\frac{O(m^{10})}{\lambda}$
and since $\lambda=\Omega(m^{13})$, $\displaystyle=\lambda^{2/3}O(m^{-2}).$

∎

###### Proof of Theorem 6.2.

We use the coupling in Lemma D.1 to generate $(X,\tilde{X})$.

$\displaystyle|\mathbb{E}\phi_{f,m;a,b}(\tilde{X})-\mathbb{E}\phi_{f,m;a,b}(X)|$
$\displaystyle\leq\mathbb{E}|\phi_{f,m;a,b}(\tilde{X})-\phi_{f,m;a,b}(X)|$
$\displaystyle\leq\int{\Gamma}|\hat{f}(\gamma)|\frac{\mathbb{E}\left|\prod_{j=1}^{3}(W_{\gamma}^{(j)}+Z_{\gamma}^{(j)})-\prod_{j=1}^{3}W_{\gamma}^{(j)})\right|}{m^{3}|\Gamma(-1/3)|^{3}}\,d\mu_{\Gamma}(\gamma)$
by the independence of the three copies
$\displaystyle\leq\int{\Gamma}|\hat{f}(\gamma)|\frac{3(\mathbb{E}|W_{\gamma}|)^{2}\mathbb{E}|Z_{\gamma}|+3\mathbb{E}|W_{\gamma}|(\mathbb{E}|Z_{\gamma}|)^{2}+(\mathbb{E}|Z_{\gamma}|)^{3}}{m^{3}|\Gamma(-1/3)|^{3}}\,d\mu_{\Gamma}(\gamma)$
by Lemma D.2 $\displaystyle\leq\int{\Gamma}|\hat{f}(\gamma)|\frac{\lambda\cdot
O(m)}{m^{3}|\Gamma(-1/3)|^{3}}\,d\mu_{\Gamma}(\gamma)$
$\displaystyle=\lambda\|\hat{f}\|_{1}O(m^{-2}).$

Then we bound the second moment.

$\displaystyle\mathbb{E}\left|\phi_{f,m;a,b}(\tilde{X})\overline{\phi_{f,m;a,b}(\tilde{X})}-\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}\right|$
$\displaystyle\leq\int{\Gamma}\int{\Gamma}|\hat{f}(\gamma)||\overline{\hat{f}(\gamma)}|\frac{\mathbb{E}\left|\prod_{j=1}^{3}(W_{\gamma}^{(j)}+Z_{\gamma}^{(j)})\overline{(W_{\gamma^{\prime}}^{(j)}+Z_{\gamma^{\prime}}^{(j)})}-\prod_{j=1}^{3}W_{\gamma}^{(j)}\overline{W_{\gamma^{\prime}}^{(j)}}\right|}{m^{6}|\Gamma(-1/3)|^{6}}\,d\mu_{\mathbbold{\Gamma}}(\gamma)d\mu_{\mathbbold{\Gamma}}(\gamma^{\prime}),$

We expand the product
$\prod_{j=1}^{3}(W_{\gamma}^{(j)}+Z_{\gamma}^{(j)})\overline{(W_{\gamma^{\prime}}^{(j)}+Z_{\gamma^{\prime}}^{(j)})}$
and cancel the term
$\prod_{j=1}^{3}W_{\gamma}^{(j)}\overline{W_{\gamma^{\prime}}^{(j)}}$. By the
independence of the three copies, we have

$\displaystyle\mathbb{E}\left|\prod_{j=1}^{3}(W_{\gamma}^{(j)}+Z_{\gamma}^{(j)})\overline{(W_{\gamma^{\prime}}^{(j)}+Z_{\gamma^{\prime}}^{(j)})}-\prod_{j=1}^{3}W_{\gamma}^{(j)}\overline{W_{\gamma^{\prime}}^{(j)}}\right|$
$\displaystyle\leq(z_{WW}+z_{ZZ}+2z_{WZ})^{3}-z_{WW}^{3},$

where $z_{WW},z_{ZZ},z_{WZ}$ are the upper bounds in Lemma D.2. Since we
assumed $a\leq m(\ln\lambda-6\ln m)$, $b\geq m(\ln\lambda+3\ln m)$ and
$\lambda=\Omega(m^{13})$, we have

$\displaystyle\mathbb{E}|W_{\gamma}||W_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{AA}=\lambda^{2/3}O(m^{2}),$
$\displaystyle\mathbb{E}|Z_{\gamma}||Z_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{BB}=\lambda^{2/3}O(m^{-2}),$
$\displaystyle\mathbb{E}|W_{\gamma}||Z_{\gamma^{\prime}}|$ $\displaystyle\leq
z_{CC}=\lambda^{2/3}O(1).$

Thus we have

$\displaystyle(z_{WW}+z_{ZZ}+2z_{WZ})^{3}-z_{WW}^{3}=\lambda^{2}O(m^{4}).$

Thus

$\displaystyle\mathbb{E}\left|\phi_{f,m;a,b}(\tilde{X})\overline{\phi_{f,m}(X)}-\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}\right|$
$\displaystyle\leq\lambda^{2}\|\hat{f}\|_{1}^{2}O(m^{-2}).$

We conclude that

$\displaystyle|\mathbb{V}(\phi_{f,m;a,b}(X))-\mathbb{V}(\phi_{f,m;a,b}(\tilde{X}))|$
$\displaystyle\leq\left|\mathbb{E}\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}-(\mathbb{E}\phi_{f,m;a,b}(X))\overline{(\mathbb{E}\phi_{f,m;a,b}(X))}\right.$
$\displaystyle\quad\quad\left.{}-\mathbb{E}\phi_{f,m;a,b}(\tilde{X})\overline{\phi_{f,m;a,b}(\tilde{X})}+(\mathbb{E}\phi_{f,m;a,b}(\tilde{X}))\overline{(\mathbb{E}\phi_{f,m;a,b}(\tilde{X}))}\right|$
Note that for two complex numbers $w_{1},w_{2}$,
$|w_{1}\overline{w_{1}}-w_{2}\overline{w_{2}}|\leq|w_{1}-w_{2}|(|w_{1}|+|w_{2}|)$.
$\displaystyle\leq\mathbb{E}\left|\phi_{f,m;a,b}(\tilde{X})\overline{\phi_{f,m;a,b}(\tilde{X})}-\phi_{f,m;a,b}(X)\overline{\phi_{f,m;a,b}(X)}\right|$
$\displaystyle\quad+\mathbb{E}|\phi_{f,m;a,b}(\tilde{X})-\phi_{f,m;a,b}(X)|(|\mathbb{E}\phi_{f,m;a,b}(\tilde{X})|+|\mathbb{E}\phi_{f,m;a,b}(X)|)$
$\displaystyle=\lambda^{2}\|\hat{f}\|_{1}^{2}O(m^{-2})+\lambda\|\hat{f}\|_{1}O(m^{-2})(\lambda|\mathbb{E}f(R)|+\lambda\|\hat{f}\|_{1}O(m^{-1}))$
where $|\mathbb{E}f(R)|\leq\|f\|_{\infty}$ and, by Lemma 2.4,
$\|f\|_{\infty}\leq\|\hat{f}\|_{1}$. In conclusion,
$\displaystyle=\lambda^{2}\|\hat{f}\|_{1}^{2}O(m^{-2}).\hskip 227.62204pt\qed$


aainstitutetext: School of Physics and Astronomy, University of Nottingham,
Nottingham NG7 2RD, UKbbinstitutetext: Nottingham Centre of Gravity,
University of Nottingham, Nottingham NG7 2RD, UK

# Dilatonic Effect in Double Field Theory Cosmology

Yang Liu<EMAIL_ADDRESS>
###### Abstract

In this article we discuss some aspects of double field theory cosmology with
an emphasis on the role played by the dilaton. The cosmological solutions of
double field theory equations of motion after coupling a shifted dilaton to
them are investigated. The equations of motion for a constant shifted dilaton
and a constant usaul dilaton in an FRW universe are obtained. The solutions of
these equations are obtained in both the supergravity frame and in the winding
frame. We also consider three possible dark energy candidates in a $4D$
universe using double field theory cosmology and find some basic conditions
which the three dark energy candidates should satisfy. We consider the results
for a more general potential of shifted dilaton as well.

##  1 Introduction

Despite its great success, General Relativity (GR) faces several shortcomings
when applied to the universe. In particular, the concepts of dark matter and
dark energy must be introduced in order to provide an explanation for the
large-scale dynamics of the universe. Furthermore, in order to solve the
horizon and the flatness problems, new dynamics, such as inflation, involving
additional degrees of freedom, which may play an important role near the
strong-coupling regime at which GR breaks down, are required [1]. In order to
make quantitative predictions concerning the early universe a consistent
quantum gravity theory needs to be developed.
At present, string theory appears to be the most promising candidate for
quantum gravity theory. It is presently being used to study cosmology in the
low energy effective supergravity (SUGRA) limit at weak coupling [2]. In GR,
the only gravitational field is the spacetime metric $g_{\mu\nu}$. A Weyl
transformation on the metric can be performed to convert it from the string
frame to the Einstein frame.
The T-duality plays an important role in string theory [3]. It states that the
physics of small compact spaces of radius $R$ is equivalent to the physics of
large compact spaces of radius $1/R$ (in string units) [3]. For example, for
strings on a torus of radius $R$, the symmetry implies that the spectrum of
string states is unchanged if $R\rightarrow 1/R$ (in string units) and string
momentum modes are interchanged with string winding modes. T-duality symmetry
is assumed to be a fundamental symmetry of non-perturbative string theory [4].
The fields of double field theory (DFT) are $O(D,D)$ generalizations of
spacetime fields [5,6,7,8,9]. The equivalence of spacetime momenta and winding
numbers in the string spectra gives rise to a set of dual coordinates
$\tilde{x}_{i}$, conjugated to winding numbers [10]. These dual coordinates
are treated on the same footing as the usual coordinates $x_{i}$. Then
spacetime dimension has changed from $D$ to $2D$. DFT is given (see, e.g.,
[11] for a review) by an action for a generalized metric in $2D$ space-time
dimensions which is constructed from the metric, the antisymmetric tensor
field and the dilaton (the “background”) of the massless sector of $D$ space-
time dimensional string theory. In particular, after imposing a section
condition the dynamical equations for the background reduce to those of
supergravity. T-duality is already present in supergravity even in the absence
of the $O(D,D)$ covariant structure introduced in DFT. However, T-duality is
manifested as an $O(D,D)$ symmetry in the action of DFT [10]. The full set of
coordinates in DFT can be denoted as $X^{M}(x^{i},\tilde{x}^{i})$, where
$x^{i}(i=1,2,...,D)$ is the usual spacetime coordinate, $\tilde{x}^{i}$ the
dual coordinate and $M=1,2,...,2D$ the $O(D,D)$ index. All of the spacetime
component fields depend on both the usual and the dual coordinates, i.e.,
$\phi_{I}=\phi_{I}(x^{i},\tilde{x}^{i})$ [10].
Cosmology in double field theory is a relatively new research field which has
been investigated only recently and at present only a few papers are available
on this topic. DFT equations of motion for cosmology has been studied in
ref.[10] but in the absence of sources of matter. The authors demonstrate that
the contraction of the conjugated space leads to both an inflation phase and a
decelerated expansion of the ordinary space during different evolution stages
[10]. Sources of matter have been included in the general equations derived in
ref.[12]. Furthermore, ref.[3] and [13] considered the T-dual cosmological
solutions. In ref.[14], Robert Brandenberger et al. made use of the T-duality
symmetry of superstring theory and of the double geometry from double field
theory and postulated that the cosmological singularities of a homogeneous and
isotropic universe would disappear. Furthermore, Peng Wang et al. considered
the effect of non-singular cosmology via $\alpha^{\prime}$ corrections [15].
In addition, in ref.[16,17], the same authors considered the $O(D,D)$ duality
string cosmology to all orders $\alpha^{\prime}$. Further details are not
listed here since these papers are not directly related to this article.
In the present article, we discuss some aspects of double field theory
cosmology with an emphasis on the role played by the dilaton. The cosmological
solutions from double field theory equations of motion after coupling a
shifted dilaton to them are discussed in both a supergravity frame and a
winding frame. In section 2 double field theory and dual cosmology are briefly
reviewed. In section 3 the equations of motion (EQM) coupling a shifted
dilaton in an FRW metric are obtained. In section 4 the solutions of the EQM
for a constant shifted dilaton in an FRW universe are considered. In section 5
the solutions for a constant usual diffeomorphic dilaton in an FRW universe
are considered. In section 6 we consider three dark energy candidates in $4D$
DFT cosmology. In section 7 we analyze a more general potential of shifted
dilaton. In section 8 conclusions and a discussion of future work are
presented.
We adopt the following notations: capital letters $M$,$N$,… represent indices
which encompass both regular and dual spacetime dimensions; small letters $i$,
$j$,… denote indices which encompass regular $D=d+1$ spacetime dimensions.
$a(t)$ is the cosmological scale factor, where $t$ is the physical time whilst
the dual time is denoted by $\tilde{t}$. The $\omega=p/\rho$ is the equation
of state, where $p$ and $\rho$ are pressure and energy density, respectively.

##  2 Short review of double field theory and dual cosmology

###  2.1 Double field theory

Current research has primarily been focused on the massless sector of closed
string spectra [10]. The DFT action includes a $D$ dimensional metric
$g_{ij}$, the anti-symmetric Kalb-Ramond field $b_{ij}$ and the dilaton field
$\phi$. The action rewrites these fields in an $O(D,D)$ covariant way, where
$D$ is the dimensionality of space-time. If the fields were only dependent on
the usual coordinates, the DFT action could be reduced to a supergravity
action [3]. The DFT action is given by [3]

$S=\int dxd\tilde{x}e^{-2d}\mathcal{R},$ (1)

where $d$ contains the usual dilaton $\phi$ and the determinant of the metric
$g$, i.e.

$e^{-2d}=\sqrt{-g}e^{-2\phi},$ (2)

and [10]

$\mathcal{R}=\frac{1}{8}\mathcal{H}^{MN}\partial_{M}\mathcal{H}^{KL}\partial_{N}\mathcal{H}_{KL}-\frac{1}{2}\mathcal{H}^{MN}\partial_{N}\mathcal{H}^{KL}\partial_{L}\mathcal{H}_{MK}-\partial_{M}d\partial_{N}\mathcal{H}^{MN}+4\mathcal{H}^{MN}\partial_{M}d\partial_{N}d,$
(3)

where the generalized metric, $\mathcal{H}_{MN}$, is defined as

$\mathcal{H}_{MN}=\begin{bmatrix}g^{ij}&-g^{ik}b_{kj}\\\
b_{ik}g^{kj}&g_{ij}-b_{ik}g^{kl}b_{lj}\end{bmatrix}.$ (4)

The level matching condition in closed string theory imposes the “weak
constraint” $\partial\tilde{\partial}\phi(x,\tilde{x})=0$ for any field
$\phi(x,\tilde{x})$. To ensure that the action is locally equivalent to the
low energy effective string action, the following so-called “strong
constraint” is required: $\partial\tilde{\partial}=0$ as an operator equation,
acting on any products of the fields [10].

###  2.2 Dual cosmology

In this article, we will set the Kalb-Ramond field $b_{ij}=0$ and adopt the
following FRW like metric:

$dS^{2}=-dt^{2}-d\tilde{t}^{2}+a^{2}(t,\tilde{t})dx^{2}+\tilde{a}^{2}(t,\tilde{t})d\tilde{x}^{2},$
(5)

where $\tilde{a}=a^{-1}$ [13].
The vacuum equations of motion for DFT in the presence of a dual time
associated with the winding sector are given by [10]

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}+4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}=0,$
(6) $(D-1)\tilde{H}^{2}-2d^{{}^{\prime\prime}}-(D-1)H^{2}+2\ddot{d}=0,$ (7)
$\tilde{H}^{{}^{\prime}}-2\tilde{H}d^{{}^{\prime}}+\dot{H}-2H\dot{d}=0,$ (8)

where a prime denotes a derivative with respect to $\tilde{t}$, and a dot
denotes a derivative with respect to $t$. Noting that $2d=2\phi-(D-1)\ln a$
defines the shifted dilaton and $\tilde{H}=a^{\prime}/a$, by variation of
action $(2.1)$ with respect to $d$, we can obtain the equation $(2.6)$ for a
shifted dilaton and by variation of action $(2.1)$ with respect to the
generalized metric $\mathcal{H}_{MN}$, we can obtain eqs.$(2.7)$ and $(2.8)$
for a graviton.
In ref.[12], the authors proposed the following cosmological equations in the
presence of matter by the following prescription:

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}+4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}=0,$
(9)
$(D-1)\tilde{H}^{2}-2d^{{}^{\prime\prime}}-(D-1)H^{2}+2\ddot{d}=\frac{1}{2}e^{2d}E(t,\tilde{t}),$
(10)
$\tilde{H}^{{}^{\prime}}-2\tilde{H}d^{{}^{\prime}}+\dot{H}-2H\dot{d}=\frac{1}{2}e^{2d}P(t,\tilde{t}),$
(11)

where $E$ and $P$ are the energy and pressure associated with the matter
sector.
In the following, we will consider the stabilized dilaton $\phi=\phi_{0}$. In
this case we have $2\dot{d}=-(D-1)H$ and $2d^{{}^{\prime}}=(D-1)\tilde{H}$,
and eqs.$(2.9)-(2.11)$ become

$2(\tilde{H}^{{}^{\prime}}-\dot{H})-D(\tilde{H}^{2}+H^{2})=0,$ (12)
$(\tilde{H}^{2}-H^{2})-(\tilde{H}^{{}^{\prime}}+\dot{H})=\frac{1}{2(D-1)}G\rho(t,\tilde{t}),$
(13)
$(\tilde{H}^{{}^{\prime}}+\dot{H})+(D-1)(H^{2}-\tilde{H}^{2})=\frac{1}{2}Gp(t,\tilde{t}),$
(14)

where $G$ depends on $\phi=\phi_{0}$ which is the fixed value of a dilaton
[3].
The mass spectrum of a closed string in one dimensional space, compactified on
a circle, is given by [3]

$M^{2}=(N+\tilde{N}-2)+p^{2}\frac{l^{2}_{s}}{R^{2}}+\omega^{2}\frac{l^{2}_{s}}{\tilde{R}^{2}},$
(15)

where $N$, $\tilde{N}$ are the oscillaratory modes of the string, $p$ denotes
the momentum modes which associated with the center of mass, and $\omega$
denotes the winding modes which represent the number of times the string has
wrapped itself around the compact dimension in a topologically nontrivial way
[3]. Taking the limit of large $R$ all physical quantities depend solely on
$t$ which identifies the so-called “supergravity frame”. Taking the limit of
small $R$, all physical quantities depend solely on $\tilde{t}$, which
identifies the so-called “winding frame” [3].

##  3 The equations of motion coupling a shifted dilaton in an FRW like
universe

Following ref.[18], we will add a dilaton potential $V(d)$ into the action
$(2.1)$, so that

$S=\int dxd\tilde{x}e^{-2d}\\{\mathcal{R}-2V(d)\\}+S_{m},$ (16)

where $S_{m}$ is the action of the matter sector. The manner of introducing
matter is described in Appendix. The reason for considering the effect of a
dilaton in double field theory cosmology is that, in standard string
cosmology, a dilaton is often regarded as quintessence (such as ref.[18]),
which can be used to explain the origin of inflation and the accelerating
expansion of the universe. In this article, we will consider whether a shifted
dilaton can be used to account for the real physical phenomena.
In order to avoid the singularity, we introduce a dilaton potential

$V(d)=V_{0}e^{8d},$ (17)

where $V_{0}>0$ [10]. It is known that non-singular solutions can be obtained
at low curvatures in the presence of an approriate non-local effective
potential [19]. This non-local potential represents the back reactions of
higher loop corrections [19]. $V_{0}$ includes a proper volume which makes the
dilaton potential a scalar under generalized diffeomorphisms. This potential
guarantees the $O(D,D)$ symmetry [10]. In fact, $V_{0}$ cannot be negative. We
should point out two points. Firstly, $V_{0}$ is a volume [14,19] which cannot
be zero. Moreover, in ref.[19], the Hubble parameter
$t^{-1}_{0}=e^{\phi_{0}}\sqrt{V_{0}}$, where $\phi_{0}$ is an integration
constant. If $V_{0}$ is negative, then the result is unphysical. However, the
prefactor of potential can be negative, i.e., the potential can be
$V(d)=-V_{0}e^{8d}$, where $V_{0}>0$. But in sections 3, 4, 5 and 6, we only
consider potential eq.$(3.2)$, which can be regarded as a particular example.
In section 7, we will consider a more general potential $V(d)=Ae^{nd}$, where
$A$ and $n$ can be arbitrary real numbers.
By variation of the shifted dilaton $d$ of action $(3.2)$, we can obtain the
equation of motion of the shifted dilaton

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}+4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}-12V_{0}e^{6d}=0.$
(18)

By variation of the generalized metric $\mathcal{H}^{MN}$ of action $(3.2)$,
the equations of motion of graviton are the same as eq.$(2.10)$ and $(2.11)$
so that the complete list of the equations of motion are:

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}+4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}-12V_{0}e^{6d}=0,$
(19)
$(D-1)\tilde{H}^{2}-2d^{{}^{\prime\prime}}-(D-1)H^{2}+2\ddot{d}=\frac{1}{2}e^{2d}E(t,\tilde{t}),$
(20)
$\tilde{H}^{{}^{\prime}}-2\tilde{H}d^{{}^{\prime}}+\dot{H}-2H\dot{d}=\frac{1}{2}e^{2d}P(t,\tilde{t}).$
(21)

In a supergravity frame, all physical quantities depend solely on physical
time $t$:

$4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}-12V_{0}e^{6d}=0,$ (22)
$-(D-1)H^{2}+2\ddot{d}=\frac{1}{2}e^{2d}E(t),$ (23)
$\dot{H}-2H\dot{d}=\frac{1}{2}e^{2d}P(t).$ (24)

In a winding frame, all physical quantities depend solely on dual time
$\tilde{t}$:

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}-12V_{0}e^{6d}=0,$
(25)
$(D-1)\tilde{H}^{2}-2d^{{}^{\prime\prime}}=\frac{1}{2}e^{2d}E(\tilde{t}),$
(26)
$\tilde{H}^{{}^{\prime}}-2\tilde{H}d^{{}^{\prime}}=\frac{1}{2}e^{2d}P(\tilde{t}).$
(27)

In the following sections, we will consider the solutions of the above
equations for constant shifted dilaton $d_{0}$ and a constant usual dilaton
$\phi_{0}$ in both a supergravity frame as well as in a winding frame.

##  4 Solutions of the equations of motion for a constant shifted dilaton in
an FRW like universe

In this section, we will obtain the solutions of the equations of motion for a
constant dilaton.
For a constant shifted dilaton $d$ in a supergravity frame, i.e., $\dot{d}=0$,
$d=d_{0}$, then the equations of motion become

$-(D-1)H^{2}-12V_{0}e^{6d_{0}}=0,$ (28)
$-(D-1)H^{2}=\frac{1}{2}e^{2d_{0}}E(t),$ (29)
$\dot{H}=\frac{1}{2}e^{2d_{0}}P(t).$ (30)

From eq.$(4.1)$, we obtain

$H=\sqrt{\frac{-12V_{0}}{D-1}}e^{3d_{0}},$ (31)

and when $V_{0}>0$, $H$ is a complex number, therefore there is no physical
solution.
In a similar manner to that presented above, we can obtain the corresponding
solutions in a winding frame:

$\tilde{H}=\sqrt{\frac{-12V_{0}}{D-1}}e^{3d_{0}},$ (32)

which is also a complex number as in the supergravity frame, therefore the
solution is unphysical.

##  5 Solutions of the equations of motion for a constant usual dilaton in an
FRW like universe

In this section, we briefly consider the T-dual solutions in both a
supergravity frame and a winding frame firstly. Then we will consider the
solutions of the equations of motion for a constant usual dilaton $\phi$ in
both a supergravity frame and in a winding frame.
For constant $\phi=\phi_{0}$, considering T-dual solutions, then we have [13]:

$2\dot{d}=\alpha(D-1)H,$ (33) $2d^{{}^{\prime}}=\tilde{\alpha}(D-1)\tilde{H},$
(34)

where $\alpha$ and $\tilde{\alpha}$ are both constants. In particular, for
$(\alpha,\tilde{\alpha})=(-1,1)$ we have a constant usual dilaton in the
supergravity frame and a non-constant usual dilaton in the winding frame; for
$(\alpha,\tilde{\alpha})=(1,-1)$ we have a constant usual dilaton in the
winding frame and nonconstant usual dilaton in the supergravity frame. The
case of $(\alpha,\tilde{\alpha})=(-1,-1)$ corresponds to having a constant
usual dilaton in both frames [13].

###  5.1 Solutions in a supergravity frame

We obtain respective sets of equations for the supergravity frame and the
winding frame. In the supergravity frame, the equations of motion are:

$2\alpha(D-1)\dot{H}-H^{2}(D-1)[\alpha^{2}(D-1)+1]-12V_{0}e^{6d}=0,$ (35)
$-(D-1)H^{2}+\alpha(D-1)\dot{H}=\frac{1}{2}e^{2d}E(t),$ (36)
$\dot{H}-\alpha(D-1)H^{2}=\frac{1}{2}e^{2d}P(t),$ (37)

therefore

$H^{2}=\frac{12V_{0}e^{6d}[\omega-\frac{1}{\alpha(D-1)}]}{[1-\alpha^{2}(D-1)][\frac{1}{\alpha}+\omega(D-1)]}.$
(38)

Applying the section conditions, then for stabilized $\phi=\phi_{0}$,
$2\dot{d}=-(D-1)H$ and $2\ddot{d}=-(D-1)\dot{H}$, then the eq.$(3.7)$ becomes

$2(D-1)\dot{H}+(D-1)^{2}H^{2}+(D-1)H^{2}+12V_{0}e^{6d}=0,$ (39)

namely,

$\dot{H}=-\frac{1}{2(D-1)}\\{H^{2}[(D-1)^{2}+(D-1)]+12V_{0}e^{6d}\\}.$ (40)

Inserting eq.$(5.8)$ into eq.$(3.8)$ and $(3.9)$, we have

$\frac{1}{2}H^{2}(D^{2}-3D+2)+6V_{0}e^{6d}=\frac{1}{2}e^{2d}E(t),$ (41)
$\frac{1}{2}H^{2}(D-2)-\frac{6V_{0}}{(D-1)}e^{6d}=\frac{1}{2}e^{2d}P(t).$ (42)

Combining eq.$(5.9)$ and $(5.10)$, we obtain that

$H^{2}=\frac{(D-1)\omega+1}{2(D-1)(D-2)}e^{2d}E(t)=\frac{(D-1)\omega+1}{2(D-1)(D-2)}e^{2\phi_{0}}a^{-(D-1)}E(t),$
(43) $E(t)=\frac{24V_{0}}{1-\omega(D-1)}e^{4\phi_{0}}a^{-2(D-1)},$ (44)

where $\omega$ is the equation of state. From eq.$(5.11)$, it is apparent $D$
cannot be 1 and 2 unless $\omega=-\frac{1}{D-1}$.
When $\omega=\frac{1}{D-1}$, $V(d)=0$, and

$H^{2}=\frac{1}{(D-1)(D-2)}e^{2\phi_{0}}a^{-(D-1)}E(t),$ (45)

which corresponds to the case with no dilaton potential [3].
In particular, for $D=4$, if $\omega=\frac{1}{D-1}=\frac{1}{3}$, where
radiation dominates the universe, then

$H^{2}=\frac{1}{6}e^{2\phi_{0}}a^{-3}E(t).$ (46)

Moreover, if we assume $\omega$ is a constant, considering eq.$(5.11)$ and
$(5.12)$, then we obtain

$a=\left(\frac{9}{4}(D-1)^{2}\right)^{\frac{1}{3(D-1)}}t^{\frac{2}{3(D-1)}},$
(47)

where we have set integration constant to be zero. Then

$\dot{a}=\left(\frac{9}{4}(D-1)^{2}\right)^{\frac{1}{3(D-1)}}\frac{2}{3(D-1)}t^{\frac{5-3D}{3(D-1)}},$
(48) $H=\dot{a}/a=\frac{2}{3(D-1)}t^{-1},$ (49)
$\ddot{a}=\left(\frac{9}{4}(D-1)^{2}\right)^{\frac{1}{3(D-1)}}\frac{10-6D}{9(D-1)^{2}}t^{\frac{8-6D}{3(D-1)}}.$
(50)

In order to account for the accelerating expansion of the universe, $\ddot{a}$
must be larger than $0$. However, this is impossible for $D\neq 1$.

###  5.2 Solutions in a winding frame

Similarly, in the winding frame, the equations of motion are:

$2\tilde{\alpha}(D-1)\tilde{H}^{{}^{\prime}}-\tilde{H}^{2}(D-1)[\tilde{\alpha}^{2}(D-1)+1]-12V_{0}e^{6d}=0,$
(51)
$(D-1)\tilde{H}^{2}-\tilde{\alpha}(D-1)\tilde{H}^{{}^{\prime}}=\frac{1}{2}e^{2d}E(\tilde{t}),$
(52)
$\tilde{H}^{{}^{\prime}}-\tilde{\alpha}(D-1)\tilde{H}^{2}=\frac{1}{2}e^{2d}P(\tilde{t}),$
(53)

therefore

$\tilde{H}^{2}=\frac{12V_{0}e^{6d}[\omega+\frac{1}{\tilde{\alpha}(D-1)}]}{[1-\tilde{\alpha}^{2}(D-1)][\omega(D-1)-\frac{1}{\tilde{\alpha}}]}.$
(54)

For stabilized $\phi=\phi_{0}$, $2d=2\phi_{0}-(D-1)\ln a$ and
$2d^{{}^{\prime}}=(D-1)\tilde{H}$, we obtain

$\tilde{H}^{2}=\frac{[(D-1)\omega+1]}{2(2-D)(D-1)}e^{2d}E(\tilde{t})=\frac{[(D-1)\omega+1]}{2(2-D)(D-1)}e^{2\phi_{0}}a^{-(D-1)}E(\tilde{t}),$
(55) $E(\tilde{t})=\frac{24V_{0}}{1-\omega(D-1)}e^{4\phi_{0}}a^{-2(D-1)},$
(56)

where $\omega$ is the equation of state. From eq.$(5.23)$, it is apparent $D$
cannot be 1 and 2 unless $\omega=-\frac{1}{D-1}$.
When $\omega=\frac{1}{D-1}$, $V(d)=0$, then

$\tilde{H}^{2}=-\frac{1}{(D-2)(D-1)}e^{2d}E(\tilde{t})=-\frac{1}{(D-2)(D-1)}e^{2\phi_{0}}a^{-(D-1)}E(\tilde{t}),$
(57)

which is the case without a dilaton potential [3].
In particular, for $D=4$, if $\omega=1/3$, then

$\tilde{H}^{2}=-\frac{1}{6}e^{2d}E(\tilde{t})=-\frac{1}{6}e^{2\phi_{0}}a^{-3}E(\tilde{t}).$
(58)

Moreover, if we assume $\omega$ is a constant, considering eq.$(5.24)$ and
$(5.25)$, then we obtain

$a=\left(\frac{9}{4}(D-1)^{2}\right)^{\frac{1}{3(D-1)}}\tilde{t}^{\frac{2}{3(D-1)}},$
(59)

where we have set integration constant to be zero. Then

$a^{\prime}=\left(\frac{9}{4}(D-1)^{2}\right)^{\frac{1}{3(D-1)}}\frac{2}{3(D-1)}\tilde{t}^{\frac{5-3D}{3(D-1)}},$
(60) $H=a^{\prime}/a=\frac{2}{3(D-1)}\tilde{t}^{-1},$ (61)
$a^{\prime\prime}=\left(\frac{9}{4}(D-1)^{2}\right)^{\frac{1}{3(D-1)}}\frac{10-6D}{9(D-1)^{2}}\tilde{t}^{\frac{8-6D}{3(D-1)}}.$
(62)

##  6 Candidates for dark energy

Since $V_{0}>0$, in order to ensure that $a$ is real, according to $(5.16)$,
we must require

$\frac{\frac{1}{D-1}+\omega}{(D-2)[1-\omega(D-1)]}>0.$ (63)

From eq.$(6.1)$, we obtain two “critical” dimensions. One is $D=2$, and the
other one is $D=1+\frac{1}{\omega}$. Therefore, for “critical” dimensions,
$\omega=\frac{1}{n}$, where $n$ denotes positive integers.
From eq.$(6.1)$, we can obtain the constraints for $\omega$ to be

$-\frac{1}{D-1}<\omega<\frac{1}{D-1}.$ (64)

Therefore, for the real universe, i.e., $D=4$ the constraints on $\omega$ are,
$-1/3<\omega<1/3$. Based on ref.[22], we know that for any general spacetime
dimension $D$, tensile matter has the equation of state

$p=\omega\rho.$ (65)

In order to explain the origin of the accelerating expansion of the universe
[22]

$-1<\omega<-(\frac{D-3}{D-1}),$ (66)

namely, $-1<\omega<-1/3$. We therefore conclude that this model with a
constant usual $\phi=\phi_{0}$ cannot be used as an explanation for the
accelerating expansion of the universe.
We will now consider three possible dark energy candidates in $D=4$ double
field theory cosmology.
Firstly, for a holographic dark energy scenario
$\rho_{D}=3c^{2}m^{2}_{p}/L^{2}$, where $c$ is a dimensionless parameter
(usually we take $c>0$), $m^{2}_{p}$ is Planck mass and $L=ar(t)=R_{h}$, where
$R_{h}$ is the radial size of the horizon [23]. If an interaction between
matter and holographic dark energy exists, i.e.

$\dot{\rho}_{D}+3H\rho_{D}(1+\omega_{D})=-Q,$ (67)
$\dot{\rho}_{m}+3H\rho_{m}=Q,$ (68)

where $Q=3b^{2}H(\rho_{m}+\rho_{D})$, $b^{2}$ is a coupling parameter and
$\rho_{m}$ and $\rho_{D}$ are the energy densities of matter and dark energy
respectively [24] then

$\omega_{D}=-\frac{1}{3}-\frac{2\sqrt{\Omega_{D}}}{3c}-\frac{b^{2}}{\Omega_{D}},$
(69)

where $\omega_{D}$ is the equation of state of dark energy, $c$ is a
dimensionless parameter and $\Omega_{D}$ is the fractional energy density
which is defined as:

$\Omega_{D}=\frac{\rho_{D}}{\rho_{cr}},$ (70)

where $\rho_{cr}=3m^{2}_{p}H^{2}$. Therefore $\omega_{D}<-\frac{1}{3}$ when
$c>0$. Considering eq.$(6.2)$, so that in view of eq.$(6.2)$ a holographic
dark energy scenario cannot exist in $D=4$ DFT cosmology.
Secondly, for a ghost dark energy scenario, $\rho_{D}=\alpha H$, where
$\alpha$ denotes a constant of order $\Lambda^{3}_{QCD}$ and QCD represents
the QCD mass scale and $H$ is the Hubble constant. Taking the same interaction
$Q$ between matter and dark energy as in eq.$(6.5)$ and $(6.6)$, we have [25]

$\omega_{D}=-\frac{1}{2-\Omega_{D}}(1+\frac{2b^{2}}{\Omega_{D}}),$ (71)

therefore

$\frac{\frac{1}{3}+\omega_{D}}{1-3\omega_{D}}=-\frac{1+\Omega_{D}+\frac{6b^{2}}{\Omega_{D}}}{3(5-\Omega_{D}+\frac{6b^{2}}{\Omega_{D}})}<0,$
(72)

since $0<\Omega_{D}<1$. According to eq.$(5.16)$, $A_{D}$ is a complex number
so that in view of eq.$(6.2)$, the ghost dark energy scenario cannot exist in
$D=4$ DFT cosmology.
Thirdly, for a Tsallis holographic dark energy scenario
$\rho_{D}=BH^{-2\delta+4}$, where $B$ is an unknown parameter and $\delta$
represents the non-additivity parameter [26]. Taking the same interaction $Q$
between matter and dark energy as in eq.$(6.5)$ and $(6.6)$, we have: [27]

$\omega_{D}=\frac{\delta-1+b^{2}/\Omega_{D}}{(2-\delta)\Omega_{D}-1}.$ (73)

Considering eq.$(6.1)$, if Tsallis holographic energy can exist in $D=4$
double field theory cosmology of the universe, then the parameters must
satisfy the inequalities:

$(2-\delta)\Omega^{2}_{D}+(3\delta-4)\Omega_{D}+3b^{2}>0,$ (74)
$(2-\delta)\Omega^{2}_{D}+(-3\delta+2)\Omega_{D}+3b^{2}>0,$ (75)

or

$(2-\delta)\Omega^{2}_{D}+(3\delta-4)\Omega_{D}+3b^{2}<0,$ (76)
$(2-\delta)\Omega^{2}_{D}+(-3\delta+2)\Omega_{D}+3b^{2}<0.$ (77)

##  7 Analysis of a more general potential

In this section, we will analyze a more general potential $V(d)=Ae^{nd}$,
where $A$ and $n$ are arbitrary real numbers. We will see how the results in
sections 3, 4 and 5 change.

###  7.1 Case 1: A>0

When $A>0$, the potential can be written as $V(d)=V_{0}e^{nd}$, where
$V_{0}>0$.
In a supergravity frame, the equations of motion are:

$4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}-2(n-2)V_{0}e^{(n-2)d}=0,$ (78)
$-(D-1)H^{2}+2\ddot{d}=\frac{1}{2}e^{2d}E(t),$ (79)
$\dot{H}-2H\dot{d}=\frac{1}{2}e^{2d}P(t).$ (80)

In a winding frame, the equations of motion are:

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}-2(n-2)V_{0}e^{(n-2)d}=0,$
(81)
$(D-1)\tilde{H}^{2}-2d^{{}^{\prime\prime}}=\frac{1}{2}e^{2d}E(\tilde{t}),$
(82)
$\tilde{H}^{{}^{\prime}}-2\tilde{H}d^{{}^{\prime}}=\frac{1}{2}e^{2d}P(\tilde{t}).$
(83)

####  7.1.1 Solutions for a constant shifted dilaton

In a supergravity frame, a constant shifted dilaton means $\dot{d}=0$ and
$d=d_{0}$. Then according to eq.$(7.1)$-$(7.3)$, we obtain 3 cases:
if $n<2$, then

$H=\sqrt{\frac{-2(n-2)V_{0}e^{(n-2)d_{0}}}{D-1}},\qquad
E=4(n-2)V_{0}e^{(n-4)d_{0}},\qquad P=0,$ (84)

which are physical solutions;
if $n=2$, then $H=0$, $E=0$ and $P=0$;
if $n>2$, then the solutions are unphysical.
In a winding frame, a constant shifted dilaton means $d^{\prime}=0$ and
$d=d_{0}$. Then according to eq.$(7.4)$-$(7.6)$, we obtain 3 cases as well:
if $n<2$, then

$\tilde{H}=\sqrt{\frac{-2(n-2)V_{0}e^{(n-2)d_{0}}}{D-1}},\qquad
E=-4(n-2)V_{0}e^{(n-4)d_{0}},\qquad P=0,$ (85)

which are physical solutions;
if $n=2$, then $H=0$, $E=0$ and $P=0$;
if $n>2$, then the solutions are unphysical.

####  7.1.2 Solutions for a constant usual dilaton

In a supergravity frame, a constant usual dilaton means $\phi=\phi_{0}$,
$2\dot{d}=-(D-1)H$ and $2\ddot{d}=-(D-1)\dot{H}$. Then according to eq.$(7.1)$
to $(7.3)$, we have

$H^{2}=\frac{1+\omega(D-1)}{2(D-1)(D-2)}e^{2d}E(t),$ (86)

and

$E(t)=\frac{4(n-2)V_{0}e^{(n-4)d}}{1-\omega(D-1)},\qquad P(t)=\omega E(t).$
(87)

In a winding frame, a constant usual dilaton means $\phi=\phi_{0}$,
$2d^{\prime}=(D-1)\tilde{H}$ and $2d^{\prime\prime}=(D-1)\tilde{H}^{\prime}$.
Then according to eq.$(7.4)$ to $(7.6)$, we have

$\tilde{H}^{2}=\frac{1+\omega(D-1)}{2(D-1)(D-2)}e^{2d}E(\tilde{t}),$ (88)

and

$E(\tilde{t})=\frac{4(n-2)V_{0}e^{(n-4)d}}{\omega(D-1)-1},\qquad
P(\tilde{t})=\omega E(\tilde{t}).$ (89)

###  7.2 Case 2: A<0

When $A<0$, we can rewrite the potential as $V(d)=-V_{0}e^{nd}$, where
$V_{0}>0$.
In a supergravity frame, the equations of motion are:

$4\ddot{d}-4\dot{d}^{2}-(D-1)H^{2}+2(n-2)V_{0}e^{(n-2)d}=0,$ (90)
$-(D-1)H^{2}+2\ddot{d}=\frac{1}{2}e^{2d}E(t),$ (91)
$\dot{H}-2H\dot{d}=\frac{1}{2}e^{2d}P(t).$ (92)

In a winding frame, the equations of motion are:

$4d^{{}^{\prime\prime}}-4d^{{}^{\prime}2}-(D-1)\tilde{H}^{2}+2(n-2)V_{0}e^{(n-2)d}=0,$
(93)
$(D-1)\tilde{H}^{2}-2d^{{}^{\prime\prime}}=\frac{1}{2}e^{2d}E(\tilde{t}),$
(94)
$\tilde{H}^{{}^{\prime}}-2\tilde{H}d^{{}^{\prime}}=\frac{1}{2}e^{2d}P(\tilde{t}).$
(95)

####  7.2.1 Solutions for a constant shifted dilaton

In a supergravity frame, a constant shifted dilaton means $\dot{d}=0$ and
$d=d_{0}$. Then according to eq.$(7.13)$-$(7.15)$, we obtain 3 cases:
if $n>2$, then

$H=\sqrt{\frac{2(n-2)V_{0}e^{(n-2)d_{0}}}{D-1}},\qquad
E=-4(n-2)V_{0}e^{(n-4)d_{0}},\qquad P=0,$ (96)

which are physical solutions;
if $n=2$, then $H=0$, $E=0$ and $P=0$;
if $n<2$, then the solutions are unphysical.
In a winding frame, a constant shifted dilaton means $d^{\prime}=0$ and
$d=d_{0}$. Then according to eq.$(7.16)$-$(7.18)$, we obtain 3 cases as well:
if $n>2$, then

$\tilde{H}=\sqrt{\frac{2(n-2)V_{0}e^{(n-2)d_{0}}}{D-1}},\qquad
E=4(n-2)V_{0}e^{(n-4)d_{0}},\qquad P=0,$ (97)

which are physical solutions;
if $n=2$, then $H=0$, $E=0$ and $P=0$;
if $n<2$, then the solutions are unphysical.

####  7.2.2 Solutions for a constant usual dilaton

In a supergravity frame, a constant usual dilaton means $\phi=\phi_{0}$,
$2\dot{d}=-(D-1)H$ and $2\ddot{d}=-(D-1)\dot{H}$. Then according to
eq.$(7.13)$ to $(7.15)$, we have

$H^{2}=\frac{1+\omega(D-1)}{2(D-1)(D-2)}e^{2d}E(t),$ (98)

and

$E(t)=\frac{-4(n-2)V_{0}e^{(n-4)d}}{1-\omega(D-1)},\qquad P(t)=\omega E(t).$
(99)

In a winding frame, a constant usual dilaton means $\phi=\phi_{0}$,
$2d^{\prime}=(D-1)\tilde{H}$ and $2d^{\prime\prime}=(D-1)\tilde{H}^{\prime}$.
Then according to eq.$(7.16)$ to $(7.18)$, we have

$\tilde{H}^{2}=-\frac{1+\omega(D-1)}{2(D-1)(D-2)}e^{2d}E(\tilde{t}),$ (100)

and

$E(\tilde{t})=\frac{-4(n-2)V_{0}e^{(n-4)d}}{\omega(D-1)-1},\qquad
P(\tilde{t})=\omega E(\tilde{t}).$ (101)

##  8 Conclusion and discussion

The T-duality plays an important role in string theory [3]. Due to the
equivalence of spacetime momenta and winding numbers in the string spectra, a
set of dual coordinates $\tilde{x}^{i}$, which is conjugated to winding
numbers [10] has been introduced in double field theory (DFT). These dual
coordinates are treated on the same footing as the usual coordinates $x^{i}$.
Therefore, the dimensionality of spacetime is from $D$ to $2D$ naturally.
Cosmology in double field theory is a relatively new research field which has
received attention only in the last few recent years. At present there are
relatively few papers dealing with this topic. The equations of motion for
double field theory cosmology have been studied in ref.[10] in the situation
where sources of matter are absent. Furthermore, the situation where sources
of matter are present has have been discussed in the general equations derived
in ref.[12]. T-dual cosmological solutions have also been dealt with in
refs.[3], [13] and [29].
In this article, we have discussed some aspects of double field theory
cosmology with an emphasis on the role played by the dilaton and have
investigated the cosmological solutions from double field theory equations of
motion after coupling a shifted dilaton to them. In fact, many researches have
been done to study the properties of dilaton potential in classical
supergravity and double field theory to higher loop orders, such as
refs.[30,31]. However, the dilatonic effect in DFT cosmology has not been
discussed. In this acticle, we only consider the effect to lowest order. The
equations of motion coupling a shifted dilaton in an FRW universe have been
obtained in section 3. In section 4, we obtained the solutions of the
equations of motion for a constant shifted dilaton in both a supergravity
frame and a winding frame. For the potential we studied in section 4, the
solutions are unphysical in both supergravity frame and winding frame. In
section 5, we briefly considered the T-dual solutions in both frames and
obtained the solutions of a constant usual diffeomorphic dilaton $\phi_{0}$ in
an FRW universe in both a supergravity frame and in a winding frame
respectively. In section 6, we found the basic conditions for three dark
energy candidates in $D=4$ DFT cosmology. In section 7, we analyzed a more
general potential $V(d)=Ae^{nd}$, where $A$ can be arbitrary real numbers.
More general results have been obtained.
A further line of study using string theory would be to use exceptional field
theory (ExFT) [11] of which Double field theory (DFT) is a subset. Both
geometrization of the $p$-form fluxes in 10-/11$d$ supergravity and their
unification with the metric of degrees of freedom would also be interesting
lines for further research. DFT corresponds to the $NS-NS$ sector of 10$D$
supergravity, whilst ExFT would also include the $R-R$ sector of 10$D$ type
$II$ supergravity or would work for 11$D$ supergravity. These theories are
sometimes referred to as being “duality invariant”: in this sense, DFT
captures T-duality and ExFT extends this to U-duality. There have been no
studies so far of cosmology using exceptional field theory as proposed above
and in future work we therefore intend to investigate the effects of ExFT
cosmology.

##  Appendix A Appendix: A note concerning T-dualizing matter

Matter is introduced in supergravity by the following action [3, 28]:

$S=\int d^{D}x\sqrt{-g}e^{-2\phi}f=\int dt\sqrt{-g_{tt}}F(\log
a,\beta\sqrt{-g_{tt}},\phi),$ (102)

where $F$ is the free energy,

$F=\int d^{D-1}xa^{D-1}e^{-2\phi}f.$ (103)

We can consider a T-dual covariant generalization of $F$ by defining the
following action:

$S=\int d^{D}xd^{D}{\tilde{x}}e^{-2\phi}\mathcal{F},$ (104)

with $\mathcal{F}$ dependent on both sets of coordinates. Then we can write

$S=\int dt\sqrt{-g_{tt}}\\{\int d^{D-1}\tilde{x}d\tilde{t}F\\}$ (105)

or

$S=\int d\tilde{t}\sqrt{-\frac{1}{g_{tt}}}\\{\int d^{D-1}\tilde{x}dtF\\},$
(106)

where $F$ is also a function of both sets of coordinates.
Finally, the standard definitions of the energy and pressure of the system are
derived as usual according to

$E(t,\tilde{t})=-2\frac{\delta F}{\delta g_{00}},$ (107)
$P_{i}(t,\tilde{t})=-\frac{\delta S}{\delta\ln a_{i}}.$ (108)

All data generated or analysed during this study are included in this
published article.

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