0$ such that \begin{equation*} \frac{q}{q-1}\frac{q^2}{q^2-1}\cdots\frac{q^n}{q^n-1}0$. Then \begin{equation*} \sum _{\lambda\in F}d_{\lambda\cup\lambda}|\phi_{\lambda}|^{2(n+c)} 0$. \end{lemma} \begin{proof} First, note that as a consequence of~\eqref{eq:1}, the simpler inequality \begin{equation*} \phi_{\lambda}\leq \frac{q^{2n-1}}{(q^{2n}-1)(q^{2n-2}-1)}\sum _{d(\varphi)=1}\sum _{(i,j)\in C(\lambda(\varphi))}\frac{(q^{2i}-1)}{(q-1)} \end{equation*} holds. Then note that $i\leq l(\lambda(\varphi))$ and so for some $\varepsilon>0$ this is bounded by \begin{equation*} \sum _{d(\varphi)=1}n\frac{q^{2l(\lambda(\varphi))}(1+\varepsilon)}{q^{2n-2}}\leq qn\frac{q^{2\max_{d(\varphi)=1} l(\lambda(\varphi))}(1+\varepsilon)}{q^{2n-2}}, \end{equation*} where $\varepsilon<1$ for large enough $n$. Now sum over $F$ to obtain \begin{equation*} \sum _{\lambda\in F}d_{\lambda\cup\lambda}(\phi_{\lambda})^{2n}\leq \sum _{d(\varphi)=1}\sum _{i=n^{0.6}}^n\sum _{l(\lambda(\varphi))=n-i}d_{\lambda\cup\lambda}\left(\frac{qn(1+\varepsilon)}{q^{2i-2}}\right)^{2n} \end{equation*} and apply Lemma~\ref{Lemma: Dimension Bound} to obtain the bound \begin{equation*} \begin{split} \sum _{\lambda\in F}d_{\lambda\cup\lambda}(\phi_{\lambda})^{2n}&\leq \sum _{d(\alpha)=1}\sum _{i=n^{0.6}}^n\frac{Cq^{2i+4in-2i^2}(n(1+\varepsilon))^{2n}}{(q^{2i-3})^{2n}} \\&\leq C(q-1)nq^{n(-2n^{0.2}+8+C'\log n)}, \end{split} \end{equation*} where $C'$ is a positive constant. For large enough $n$ this is exponentially small. \end{proof} Finally, bound the remaining case of large length. \begin{lemma} \label{lemma:large length} If $F$ denotes the set of partition-valued functions $\lambda$ such that $\phi_{\lambda}>0$ and there is some $\varphi$ with $d(\varphi)=1$ and $l(\lambda(\varphi))>n-n^{0.6}$, then \begin{equation*} \sum _{\lambda\in F}d_{\lambda\cup\lambda}|\phi_{\lambda}|^{2(n+c)}\leq Ae^{-Bc} \end{equation*} for constants $A,B>0$ and sufficiently large $n$. \end{lemma} \begin{proof} Consider Equation~\eqref{eq:1}, first bounding the term $(n-i,1)$ in $C(\lambda(\varphi))$ corresponding to the $\varphi$ witnessing $\lambda(\varphi)>n^{0.6}$, so $i\leq n^{0.6}$. This gives \begin{equation*} \frac{q^{2n-2}(q^{2n-2i}-1)}{(q^{2n}-1)(q^{2n-2}-1)}\leq \frac{1+q^{-n}}{q^{2i}} \end{equation*} for large enough $n$. Now if $D$ denotes all other terms in the sum over $\varphi$ and $C(\lambda(\varphi))$ except for this $(n-i,1)$ term, then \begin{equation*} \sum _{D}\frac{(q^j-1)(q^{2i}-1)}{q^{j-1}(q-1)(q^2-1)}\leq q^{2n^{0.6}} \end{equation*} because the $q^{2i}$ in the remaining summands have $\sum i\leq n^{0.6}$ and so the sum is at most $q^{2n^{0.6}}$ since $q^x+q^y\leq q^{x+y}$ for $x,y\geq 1$. Then for large enough $n$, \begin{equation*} \frac{q^{2n-2}(q^2-1)}{(q^{2n}-1)(q^{2n-2}-1)}q^{2n^{0.6}}\leq \frac{q^{-n}}{q^{2i}} \end{equation*} and so \begin{equation*} \phi_{\lambda}\leq \frac{1+2q^{-n}}{q^{2i}}. \end{equation*} Now, use Lemma~\ref{Lemma: Dimension Bound} to conclude \begin{equation*} \begin{aligned} \sum _{d(\varphi)=1}\sum _{i=1}^{n^{0.6}}\sum _{l(\lambda(\varphi))=n-i}d_{\lambda\cup\lambda}\phi_{\lambda}^{2(n+c)} %\\ &\leq \sum _{d(\varphi)=1}\sum _{i=1}^{n^{0.6}} Cq^{2i+4in-2i^2}(1+2q^{-n})^{2(n+c)}q^{-4i(n+c)} \\&=(q-1)\sum _{i=1}^{n^{0.6}} C(1+2q^{-n})^{2(n+c)}q^{2i-2i^2-4ic} \\&\leq \sum _{i=1}^{n^{0.6}} C(1+2q^{-n})^{2(n+c)}q^{3-2i-4ic}. \end{aligned} \end{equation*} Now note that $(1+2q^{-n})^{2n}\rightarrow 1$ as $n\rightarrow \infty$ and so for large enough $n$ \begin{equation*} \begin{split} \sum _{d(\varphi)=1}\sum _{i=1}^{n^{0.6}}\sum _{l(\lambda(\varphi))=n-i}d_{\lambda\cup\lambda}\phi_{\lambda}^{2(n+c)}&\leq C\sum _{i=1}^{n^{0.6}}(1+2q^{-n})^{2c}q^{3-2i-4ic} \\&\leq Cq^3\sum _{i=1}^{n^{0.6}}(q^{-2-3c})^i \\&\leq Cq^{1-3c} \end{split} \end{equation*} and this establishes the lemma. \end{proof} Finally, Theorem~\ref{thm:upper bound} follows easily from Lemmas~\ref{lemma: negative case}, \ref{Lemma:Small Case} and~\ref{lemma:large length}, noting that these lemmas cover all non-zero terms in Equation~\eqref{eq: DS bound}. \section{Lower Bounds} \label{sec: lower bound} This section is devoted to proving Theorem~\ref{thm:lower bound}. This is done by showing that if only $n-c$ steps are taken, then the proportion of $\GL_{2n}(\mathbf{F}_q)$ which can be reached is exponentially small, and so the walk cannot be mixed. Similar ideas appear in~\cite{H92}, where it is shown that for the random walk on $\GL_n(\mathbf{F}_q)$, after taking $n-c$ steps the random element still has a large fixed subspace. This relies on results in an unpublished manuscript of Rudvalis and Shinoda about the proportion of such elements in $\GL_n(\mathbf{F}_q)$. It will be shown that if only $n-c$ steps are taken, then $g^TJg-J$ has an isotropic subspace of dimension $n+c$ (that is, a subspace where the form restricts to $0$). This is done by using results in~\cite{BKS90} to reduce the problem to the computation in the $\GL_n(\mathbf{F}_q)$ case. \begin{prop} \label{prop:supp estimate} Let \begin{equation*} A_c=\{g\in \GL_{2n}(\mathbf{F}_q)|g=kg_0,\dim(\ker(g_0-I))\geq n+c, k\in \Sp_{2n}(\mathbf{F}_q)\}. \end{equation*} Then \begin{equation*} \frac{|A_c|}{|\GL_{2n}(\mathbf{F}_q)|}\leq \frac{C}{q^c} \end{equation*} for some constant $C$ independent of $n$. \end{prop} \begin{proof} Note that if $g\in A_c$, then $g^TJg-J=g_0^TJg_0-J$ is an alternating form with an isotropic subspace of dimension $n+c$, because $g_0$ fixes a subspace of dimension $n+c$. Now, if $I(\omega)$ denotes a maximal isotropic subspace of the form $\omega$, then \begin{equation*} \begin{split} |A_c|&\leq \sum _{\dim(I(g_\mu^TJg_\mu-J))\geq n+c}|Kg_\mu K| \\&=\sum _{\dim(\ker(g_\mu-I))\geq n+c}|Kg_\mu K| \end{split} \end{equation*} because the set can be broken up into double cosets as it is $\Sp_{2n}(\mathbf{F}_q)$-bi-invariant since \begin{equation*} \dim(I(g_\mu^TJg_\mu-J))=\dim(I((k_1g_\mu k_2)^TJk_1g_\mu k_2-J)). \end{equation*} Thus the dimension of the maximal isotropic subspace of \begin{equation*} \begin{split} g_\mu^TJg_\mu-J&=\left(\begin{matrix} M_\mu^T&0\\ 0&I\\ \end{matrix}\right) \left(\begin{matrix} 0&I\\ -I&0\\ \end{matrix}\right) \left(\begin{matrix} M_\mu&0\\ 0&I\\ \end{matrix}\right)-\left(\begin{matrix} 0&I\\ -I&0\\ \end{matrix}\right) \\&=\left(\begin{matrix} 0&M_\mu^T-I\\ I-M_\mu&0\\ \end{matrix}\right) \end{split} \end{equation*} is just \begin{equation*} 2\dim(\ker(M_\mu-I))+\frac{1}{2}(2n-2\dim(\ker(M_\mu-I)))=n+\dim(\ker(M_\mu-I)) \end{equation*} and so $\dim(I(g_\mu^TJg_\mu-J))\geq n+c$ is equivalent to $\dim(\ker(M_\mu-I))\geq c$. But $|Kg_\mu K|=|\Sp_{2n}(\mathbf{F}_q)||C_\mu|_{q\mapsto q^2}$ by Proposition~\ref{prop: double coset size} and so \begin{equation*} |A_c|=|\Sp_{2n}(\mathbf{F}_q)| \sum _{\dim(\ker(M_\mu-I))\geq c}|C_{\mu}|_{q\mapsto q^2}. \end{equation*} Next, write \begin{equation*} |C_{\mu}|_{q\mapsto q^2}= |C_{\mu}|\frac{|C_{\mu}|_{q\mapsto q^2}}{|C_{\mu}|} \end{equation*} and note that \begin{equation*} \begin{split} \frac{|C_{\mu}|_{q\mapsto q^2}}{|C_{\mu}|}&=\frac{|\GL_n(\mathbf{F}_q)|_{q\mapsto q^2}}{|\GL_n(\mathbf{F}_q)|}\frac{a_\mu(q)}{a_\mu(q^2)} \\&= \frac{|\GL_n(\mathbf{F}_q)|_{q\mapsto q^2}}{|\GL_n(\mathbf{F}_q)|}\frac{q^{n+2n(\mu)}\prod \psi_{m_i(\mu)}(q^{-1})}{q^{2n+4n(\mu)}\prod \psi_{m_i(\mu)}(q^{-2})} \\&\leq \frac{|\GL_n(\mathbf{F}_q)|_{q\mapsto q^2}}{|\GL_n(\mathbf{F}_q)|}\frac{1}{q^n\prod_{j} \prod_{i=1}^{m_j(\mu)} (1+q^{-i})} \\&\leq \frac{|\GL_n(\mathbf{F}_q)|_{q\mapsto q^2}}{|\GL_n(\mathbf{F}_q)|}q^{-n}. \end{split} \end{equation*} Then \begin{multline*} %\begin{split} %& |\Sp_{2n}(\mathbf{F}_q)| \sum _{\dim(\ker(M_\mu-I))\geq c}|C_{\mu}|_{q\mapsto q^2} \\ \leq |\Sp_{2n}(\mathbf{F}_q)|\frac{|\GL_n(\mathbf{F}_q)|_{q\mapsto q^2}}{|\GL_n(\mathbf{F}_q)|}q^{-n}\sum _{\dim(\ker(M_\mu-I))\geq c}|C_{\mu}|. %\end{split} \end{multline*} Now from~\cite[\S~6]{H92} \begin{equation*} \sum _{\dim(\ker(M_\mu-I))\geq c}|C_{\mu}|\leq \frac{4}{q^c}|\GL_n(\mathbf{F}_q)| \end{equation*} and note that \begin{equation*} \begin{split} \frac{|\Sp_{2n}(\mathbf{F}_q)||\GL_n(\mathbf{F}_q)|_{q\mapsto q^2}q^{-n}}{|\GL_{2n}(\mathbf{F}_q)|}&= \frac{q^{n^2}\prod _{i=1}^n(q^{2i}-1)q^{n^2-n}\prod _{i=1}^n(q^{2i}-1)}{q^nq^{n(2n-1)}\prod _{i=1}^{2n}(q^i-1)} \\&=\prod _{i=1}^n\frac{q^{2i}-1}{q^{2i}-q} \end{split} \end{equation*} and this is bounded (independent of $n$ and $q$) by some $C$. Thus, \begin{equation*} \frac{|A_c|}{|\GL_{2n}(\mathbf{F}_q)|}\leq \frac{4C}{q^c}.\qedhere \end{equation*} \end{proof} Then Theorem~\ref{thm:lower bound} follows easily because $P^{\ast(n-c)}$ is supported on $A_c$. \begin{proof}[Proof of Theorem~\ref{thm:lower bound}] First, note that $P^{\ast(n-c)}$ is supported on $A_c$ because the random element may be written as \begin{equation*} k_0g_\mu k_1\cdots k_{n-1}g_\mu k_n =k_0'\prod_{i=1}^{n-c}\left((k_i')^{-1} g_\mu k_i'\right), \end{equation*} where $k_n'=k_n$, and $k_i'=k_ik_{i+1}'$, and note that each $(k_i')^{-1} g_\mu k_i'$ is a transvection, and so $\prod_{i=1}^{n-c} (k_i')^{-1} g_\mu k_i'$ is the product of $n-c$ transvections, and thus has an $(n+c)$-dimensional $1$-eigenspace. Thus, $P^{\ast(n-c)}$ is supported on $A_c$. Then as the support of $D$ has only $q$ elements, \begin{equation*} \begin{split} \|P^{\ast(n-c)}\ast D-U\|&\geq |P^{\ast(n-c)}\ast D(A_c\cdot \supp (D))-U(A_c\cdot \supp (D))| \\&\geq 1-\frac{q|A_c|}{|\GL_{2n}(\mathbf{F}_q)|} \\&\geq 1-\frac{C}{q^{c-1}} \end{split} \end{equation*} by Proposition~\ref{prop:supp estimate}. \end{proof} \bibliographystyle{amsplain-ac} \bibliography{ALCO_He_387} \end{document}