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Let $G(r,n-r)$ be the Grassmannian, which can be identified with the space of all rank $r$ projection matrices in $\mathbb{R}^n$. Let $\mu$ be the uniform measure on $G(r,n-r)$. For any $\lambda_1,...\lambda_r>0$ and $\sum_{i=1}^r\lambda_i=1$, we can define a Pseudo distance metric $d_\Lambda$ on $G(r,n-r)$: $$d_\Lambda(P,Q)=\sqrt{trace\left\{(P-Q)\Lambda(P-Q)\right\}}$$ where $$\Lambda=diag\{\lambda_1,...,\lambda_r,0,0,...,0\}$$

Define the Balls in $G(r,n-r)$ centered at $I_{r,n-r}$ with respect to $d_\Lambda$ with radius $a$ as: $$B_a(\Lambda)=\{P\in G(r,n-r): d_\Lambda(P,I_{r,n-r})\leq a\}$$ where $I_{r,n-r}$ is the diagonal matrix putting $r$ 1s on the upper left corner and $n-r$ 0s on the bottom right corner. Note that $d_\Lambda(P,I_{r,n-r})=0$ implies $P=I_{r,n-r}$.

I'm trying to prove the following conjecture: for any $1>a>b>0, a/b>100$, there exists a constant $C>0$, independent of $\Lambda$, such that $$\frac{\mu(B_a(\Lambda))}{\mu(B_b(\Lambda))}\leq\left(\frac{Ca}{b}\right)^{r(n-r)}$$

I'm trying to find a way to use Bishop-Gromov inequality. However, $d_\Lambda$ is not the geodesic distance metric. This is the tricky part. If we upper bound the nominator and lower bound the denominator separately by geodesic balls on the Grassmannian, the constant $C$ will depend on $\Lambda$, which is not desirable. This is like bounding the volume ratio of ellipsoids, the ratio should not depend on the length of axis.

Update: I have a very naive idea and do not know if it will lead to anything (with high probability nothing, I'm really new to this field). Construct n dimensional diagonal matrix $\tilde{\Lambda}=diag\{\lambda_1,...,\lambda_r,b^2/r,...,b^2/r\}$. This defines a new metric on the space of all symmetric matrices: $<A,B>=trace(A\tilde{\Lambda}B)$. Let $\tilde{\mu}$ be the new volume measure on the Grassmannian with metric induced from the new metric defined above. Then: $$\frac{\mu(B_a(\Lambda))}{\mu(B_b(\Lambda))}\leq\frac{\mu(B_{a+b}(\tilde{\Lambda}))}{\mu(B_b(\tilde{\Lambda}))}\leq\frac{\mu(B_{1.01a}(\tilde{\Lambda}))}{\mu(B_b(\tilde{\Lambda}))}$$ Now the questions is to relate $\mu$ with $\tilde{\mu}$ and relate $B_{1.01a}(\tilde{\Lambda}),B_b(\tilde{\Lambda})$ with the geodesic ball on the Grassmannian with the newly defined metric. Then we can apply Bishop-Gromov. However, I do not know if this is doable...

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$\DeclareMathOperator{\Gr}{Gr}$ It is convenient to identify $G(r,n-r)$ with the space $\newcommand{\bR}{\mathbb{R}}$ $\Gr_r(\bR^n)$ of $r$-dimensional subspaces of $\bR^n$ via the map $$ \Gr_r(\bR^n)\ni L\mapsto P_L \in G(r,n-r), $$ where $P_L$ is the orthogonal projection on $L$. The operator $I_{r,n-r}$ corresponds to the subspace $L_0=\bR^r\oplus 0\subset \bR^n$.

Denote by $\newcommand{\eO}{\mathscr{O}}$ $\eO$ the open subset of $\Gr_r(\bR^n)$ consisting of subspaces that intersect $L_0^\perp=0\oplus \bR^{n-r}$ transversally. $\DeclareMathOperator{\Hom}{Hom}$ The open set $\eO$ can be identified with $\Hom(\bR^r,\bR^{n-r})$ via the graph map $$ \Hom(\bR^r, \bR^{n-r})\ni S\mapsto \Gamma_S\in \eO, $$ $$ \Gamma_S=\big\{(x,Sx)\in\bR^r\times \bR^{n-r}:\;\;x\in\bR^r\big\}. $$ We think of matrices $S\in\Hom(\bR^r,\bR^{n-r})$ as defining coordinates on $\eO$

Using the decomposition $\bR^n=\bR^r\times\bR^{n-r}$ we can describe an $n\times n$ matrix $X$in blocks $$ X=\left[ \begin{array}{cc} A & B\\ C & D \end{array} \right], $$ where $A$ is $r\times r$, $D$ is $(n-r)\times (n-r)$ etc.

For $S\in\Hom(\bR^r,\bR^{n-r})$ (or equivalently a $(n-r)\times r$ matrix) the orthogonal projection onto $\Gamma_S$ has the block decomposition (see Eq. (1.2.5) p.17 of this book) $$ P_{\Gamma_S}=\left[ \begin{array}{cc} (1+S^*S)^{-1} & (1+S^*S)^{-1}S^*\\ S (1+S^*S)^{-1} & S (1+S^*S)^{-1} S^* \end{array} \right]. $$ For simplicity we set $W(S):=(1+S^*S)^{-1}\in\Hom(\bR^r,\bR^r)$. The plane $L_0=\bR^r\oplus 0$ has coordinate $S_0=0$ and $P_{L_0}=I_{r, n-r}$. For $L\in \eO$ with coordinates $S=S(L)$ we have $$ P_L-P_{L_0}= . \left[ \begin{array}{cc} W(S)-1 & W(S)S^*\\ S W(S) & S W(S) S^* \end{array} \right] $$ If we denote by $d_\Lambda$ the metric you defined. Then, setting $W=W(S)$, we get$\DeclareMathOperator{\tr}{tr}$ $$ d_\Lambda(L,L_0)^2=\tr (W-1)\Lambda(W-1) +\tr (SW\Lambda WS^*) $$ $$ =\tr(\Lambda(W-1)^2)+\tr(\Lambda WS^*SW) $$ ($S^*S$ commutes with $W(S)$) $$ =\tr(\Lambda(W-1)^2)+\tr(\Lambda W^2S^*S) $$ $$ =\tr\Lambda \big(\; (W-1)^2+W^2 S^*S\;\big) $$ For simplicity we set $R=R(S)=S^*S$ Thus $$ d_\Lambda(L,L_0)^2=\tr\Lambda \big(\; (W-1)^2+W^2 R\;\big). $$ Note that $$ W-1=R(1+R)^{-1}=WR,\;\;(W-1)^2=W^2R^2, $$ $$(W-1)^2+W^2R=W^2(1+R)R=WR. $$ Hence $$ d_\Lambda(L,L_0)^2=\tr\Lambda R W=\tr \Lambda R \big(1+R\big)^{-1},\;\;R=R\big(S(L)). $$ Let us see how this looks, when $r=m$, $n=n+1$. Then $S$ is a $1\times m$ matrix $$ S=(s_1,\dotsc,s_m) $$ Then $$ R(1+R)^{-1}= R- R^2+ O(\Vert R\Vert^3) $$ Denote by $E_a(\Lambda)$ the ellipsoid $$ \big\{S\in\bR^m;\;\;\tr \Lambda R(S)\leq a^2\big\}=\big\{S\in\bR^m:\;\;\lambda_1s_1^2+\cdots+\lambda_ms_m^2\leq a^2\big\}. $$ Then $$ \lim_{a\searrow 0}\frac{\mu\big( B_a(\Lambda)\big)}{ \mu\big( E_a(\Lambda)\big)}=1,\;\;\forall \Lambda,\;\;\prod_i\lambda_i>0. \tag{1} $$ The proof uses the change in coordinates $$ S=T\Lambda^{-\frac{1}{2}} ,\;\;T=(t_1,\dotsc, t_m), $$ so $$ R(S)= \Lambda^{-\frac{1}{2}}R(T)\Lambda^{-\frac{1}{2}}. $$ If the convergence in (1) were uniform in $\Lambda$, it would confirm your claim in the case $r=n-1$ and supports your initial intuitive comparison with ellipsoids.

Perhaps it would be good to look at the case $m=2$ first. In this case $$ R(S)=\left[ \begin{array}{cc}s_1^2 & s_1s_2\\ s_1s_2 & s_2^2 \end{array} \right],\;\; W(S)=\frac{1}{1+s_1^2+s^2} \left[\begin{array}{cc}1+s_2^2 & -s_1s_2\\ -s_1s_2 & 1+s_1^2 \end{array} \right]. $$ $$ RW= \frac{1}{1+s_1^2+s^2} \left[\begin{array}{cc} s_1^2 &\ast\\ \ast & s_2^2 \end{array} \right] $$ $$ \tr(\Lambda RW)=\frac{\lambda_1s_1^2+\lambda_2s_2^2}{1+s_1^2+s^2},\;\;\lambda_1+\lambda_2=1. $$ As $\lambda_1\searrow 0$ the ball $B_a(\Lambda)$ becomes unbounded in the $S$ coordinates and $\mu$ can no longer be approximated by the Euclidean volume.

The precise form of the invariant volume on $\Gr_r(\bR^m)$ is described in Sec. 9.1.2 of the same notes linked above. The computations in the case $r=2$, $n=3$ are eminently doable and you can test your hypothesis in this case.

Remark Suppose that $m=2$ and $a^2=\lambda_1$. Here $a$ is meant to be small. Then $$ B_a(\Lambda)=\big\{ (a_1,s_2)\in\bR^2;\;\;(1-a^2)s_2^2\leq a^2\big\} $$ so that in the $(s_1,s_2)$ coordinates $B_a(\Lambda)$ is, up to a $\mu$-negligible set, the strip $$ |s_2|\leq \frac{a}{\sqrt{1-a^2}}. $$ For $b^2=1-a^2$ the region $B_b(\Lambda)\cap \eO$, $\lambda_1=a^2$ has the description $$ a^2s_1^2+(1-a^2)s_2^2)\leq (1-a^2)(1+s_1^2+s_2^2) $$ i.e., $$ (1-a^2)+(1-2a^2)s_1^2\geq 0. $$ Thus $B_b(\Lambda\cap \eO=\eO$ so $$ \mu(B_b(\Lambda))= \mu(B_b(\Lambda)\cap \eO)= \mu(\eO)=\mu(\Gr_m(\bR^{m+1}). $$ If your estimate were to hold uniformly we would have $$ \mu(B_{\sqrt{\lambda_1}}(\Lambda))=O(\lambda_1). $$ Given that $B_{\sqrt{\lambda_1}}(\Lambda)$ is the strip $|s_1|\leq \sqrt{\lambda_1}$ its volume is $\geq const.\sqrt{\lambda_1})$.

To see this consider the thin rectangle $$ D_{\lambda_1}=\{|s_1|\leq 1,\;\;|s_2|\leq \sqrt{\lambda_1}\}. $$ We write $\mu$ ast $$ \mu=\rho(s_1,s_2) $$ and we observe that there exists $C_\mu>0$, independent of $\Lambda$ such that $$ \rho(s_1,s_2)\geq C_\mu,\;\;\forall |s_1|,|s_2|\leq 1. $$ Then $$ \mu(B_{\sqrt{\lambda_1}}(\Lambda))\geq \mu(D_{\lambda_1})\geq C_\mu\sqrt{\lambda_1}. $$

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  • $\begingroup$ Thank you for your brilliant answer and your notes! @Liviu Nicolaescu. You basically find a local coordinate near $I_{r,n-r}$ and approximate the volume by the volume in the local chart. A quick question is that does the set $\mathbb{O}$ contain every element in $B_a(\Lambda)$ even when $a$ is close to 1? $\endgroup$ Apr 22, 2019 at 20:28
  • $\begingroup$ What's missing is of measure zero because the complement of $\mathscr{O}$ has $\mu$-measure zero. $\endgroup$ Apr 22, 2019 at 20:31
  • $\begingroup$ Also, I do not believe the uniformity of the estimate. See Remark 1 that I've just added to my answer. $\endgroup$ Apr 22, 2019 at 21:02
  • $\begingroup$ For the remark, $b^2=1-a^2$ is not considered in my conjecture since I'm only considering only $a>b$. @Liviu Nicolaescu $\endgroup$ Apr 22, 2019 at 21:40
  • $\begingroup$ Also, I feel that only working within the local coordinates to approximate the volume near $I_{r,n-r}$ is not enough. We have to find a way to bridge the volume in the local coordinates to the one on the manifold. Is it possible to construct a new metric on the grassmannian such that the geodesic ball under this new metric around $I_{r,n-r}$ is exactly the one defined from the local coordinates so that we can use the positive curvature of grassmannian and apply Bishop-Gromov inequality? $\endgroup$ Apr 23, 2019 at 1:18

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