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...


$\DeclareMathOperator{\Gr}{Gr}$ 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 $$ \mu(B_{\sqrt{\lambda_1}}(\Lambda))\geq \mu(D_{\lambda_1})\geq C_\mu\sqrt{\lambda_1}. $$

  • $\begingroup$ 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? $\endgroup$ – neverevernever Apr 22 at 20:28
  • $\begingroup$ What's missing is of measure zero because the complement of $\mathscr{O}$ has $\mu$-measure zero. $\endgroup$ – Liviu Nicolaescu Apr 22 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. $\endgroup$ – Liviu Nicolaescu Apr 22 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 $\endgroup$ – neverevernever Apr 22 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? $\endgroup$ – neverevernever Apr 23 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.