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 and coauthors.

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 the 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 $1 - \gamma (S_d^{+})$.

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 $1 - \gamma (S_d^{+})$.

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 et. al.

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 the Nazarov's inequality, for any convex set $A$ containing the origin, it holds that \begin{align} \int_{\partial A} \frac{1}{1 + h(x) }\cdot \phi (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 $[2]$, conclude that the Gaussian surface area of the PSD cone is no larger than $1 - \gamma (S_d^{+})$.

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 and coauthors.

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 the 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 $1 - \gamma (S_d^{+})$.