Is $\{X\in\mathbb{R}^{n\times p}|X^TX=I, XX^T\circ I=\dfrac{p}{n}I\}$ a submanifold of $\mathbb{R}^{n\times p}$?

Question

$\{X\in\mathbb{R}^{n\times p}|X^TX=I\}$ is a Stiefel manifold and this is known as a submanifold of $\mathbb{R}^{n\times p}$. However, if we add the additional condition $XX^T\circ I=\dfrac{p}{n}I$, where $\circ$ is the Hadamard product, then how can we know whether $\{X\in\mathbb{R}^{n\times p}|X^TX=I, XX^T\circ I=\dfrac{p}{n}I\}$ is still a submanifold of $\mathbb{R}^{n\times p}$ or not? I would appreciate it if you could prove it, give me a hint, or explain why it's true or not.

Answer 1

This is only the beginning of an answer.

Let

$$\mathcal{S}(p,n):=\{X\in\mathbb{R}^{n\times p}|X^TX=I, XX^T\circ I=\dfrac{p}{n}I\}$$

and define the map $F:\mathbb{R}^{n\times p}\mapsto\mathbb{S}^{p}\times\mathbb{D}^{n}$ as

$$F(X)=\begin{bmatrix}X^TX-I & 0\\0 & \mathrm{diag}_i(e_i^TXX^Te_i-\frac{p}{n})\end{bmatrix}$$

where $\mathbb{S}^{p}$ is the set of real symmetric matrices of dimension $p$ and $\mathbb{D}^{n}$ is the set of real diagonal matrices of dimension $n$. Clearly, we have that $\mathcal{S}(p,n)=F^{-1}(0_{p+n})$. We now need to show that the map $F$ is a submersion at every point $X$ in $\mathcal{S}(p,n)$. In this specific case, this means that we need to show that for all $S\in\mathbb{S}^{p}\times\mathbb{D}^{n}$, there exists a $U\in\mathbb{R}^{n\times p}$ such that $DF(X)[U]=S$ where

$$DF(X)[U]=\begin{bmatrix}X^TU+U^TX & 0\\0 & \mathrm{diag}_i(e_i^T(XU^T+UX^T)e_i)\end{bmatrix}.$$

We now need to construct such a $U$, which is something I have started to think about and will complete when I have figured out if this is possible. Any comment is welcome.