Gaussian Surface Area of Positive Semidefinite Cone Let $\mathbb{R}^n$ be the Euclidean space and $A \subseteq \mathbb{R}^n$ be a sufficiently regular set, e.g., one that has smooth boundary or is convex. We define the $\epsilon$-neighbor of $A$ in the $\ell_2$ sense as
\begin{align}
A^{\epsilon} = \{ y \in \mathbb{R}^n \colon \text{there exists}~ x \in A ~\text{such that}~ \| x - y \|_{2} \leq \epsilon \}.
\end{align}
 We denote $\gamma$ as the  standard $n$-dimensional Gaussian measure $N(0, I_n)$, then the Gaussian surface measure is defined as 
\begin{align}
\tau (A) = \lim_{\epsilon \rightarrow 0} \frac{ \gamma( A^{\epsilon} \setminus A) }{  \epsilon}.
\end{align}
By results of Keith Ball (the reverse isoperimetric problem for Gaussian measure https://link.springer.com/article/10.1007/BF02573986), there is a universal upper bound of the Gaussian surface area of any convex set. 
My questions are: 1) when is Keith Ball's upper bound tight? 2) what is a tight upper bound of the Gaussian surface area of the cone of positive semidefinite matrices?
 A: The Gaussian surface area of the positive semidefinite cone $\mathbb{S}_d^{+} = \{ M \in \mathbb{R}^{d\times d} \colon M \succeq 0 \}$ is bounded by a constant in $[0,1]$ uniformly, according to ''Learning Geometric Concepts via Gaussian Surface Area'' by Adam R. Klivans, Ryan O'Donnell and Rocco A. Servedio.
More specifically, as shown in Definition 2 of this paper, the Gaussian surface area of any convex set $A\in \mathbb{R} ^{d\times d} $ can be written as 
\begin{align}
\Gamma(A) = \int_{\partial A} \varphi(x) d \sigma (x),~~~~[1]
\end{align}
where $\varphi$ is the Gaussian measure on $\mathbb{R}^{d\times d}$ and $\sigma$ is the Euclidean surface area.
By Nazarov's inequality, for any convex set $A$ containing the origin,   it holds that 
\begin{align}
\int_{\partial A} \frac{\phi (x)}{1 + h(x) } d \sigma (x) \leq 1 - \gamma (A) \leq 1, ~~~~[2]
\end{align}
where $\gamma $ is the Gaussian measure and function $h$ is defined as the distance from the origin to the tangent hyperplane of $A$ which contains $h\in \partial A$. In more details, for any $h \in \partial A$, let $n_y$ be the unit normal vector to $\partial A$ at $y$. Then  we define $h$ by $h(y)= \| y\|_2 \cdot \cos ( \langle y, n_y \rangle )$. 
Now we go back to the PSD cone. The boundary of $\mathbb{S}_d^{+}$ is defined as 
\begin{align}
\partial \mathbb{S}_d^{+} = \{ M\colon M \succeq 0, \textrm{rank}(M) < d\}.
\end{align}
For any $M \in \partial \mathbb{S}_d^{+}$, the tangent plane at $M$ is given by 
\begin{align}
\mathcal{T}(M) = \left \{ X M + MX^{\top} \colon X \in \mathbb{R}^{d\times d} \right \},
\end{align}
which implies that $M\in \mathcal{T} (M)$. Thus we have $h(M) = 0$. Combining $[1]$ and $[2]$, we conclude that the Gaussian surface area of the PSD cone is no larger than $1 - \gamma (S_d^{+})$.
A: Regarding the tightness of Ball's bound, which is $O(n^{1/4})$ for a convex set in dimension $n$, this was proven to be sharp (up to a constant factor) by Nazarov, in 
Nazarov, Fedor, On the maximal perimeter of a convex set in $\mathbb{R}^n$ with respect to a Gaussian measure, Milman, V. D. (ed.) et al., Geometric aspects of functional analysis. Proceedings of the Israel seminar (GAFA) 2001--2002. Berlin: Springer (ISBN 3-540-00485-8/pbk). Lect. Notes Math. 1807, 169-187 (2003). ZBL1036.52014.
