# Volume comparison on Grassmannian

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: $$=trace(A\tilde{\Lambda}B)$$. 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 $$\mu(B_{\sqrt{\lambda_1}}(\Lambda))\geq \mu(D_{\lambda_1})\geq C_\mu\sqrt{\lambda_1}.$$

• 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? – neverevernever Apr 22 '19 at 20:28
• What's missing is of measure zero because the complement of $\mathscr{O}$ has $\mu$-measure zero. – Liviu Nicolaescu Apr 22 '19 at 20:31
• Also, I do not believe the uniformity of the estimate. See Remark 1 that I've just added to my answer. – Liviu Nicolaescu Apr 22 '19 at 21:02
• For the remark, $b^2=1-a^2$ is not considered in my conjecture since I'm only considering only $a>b$. @Liviu Nicolaescu – neverevernever Apr 22 '19 at 21:40
• 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? – neverevernever Apr 23 '19 at 1:18