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I'm trying to find the geodesic that connects the identity with some circulant, symmetric matrix $U\in\mathrm{GL}(N,\mathbb{R})$, meaning we have \begin{align} U=\left(\begin{array}{ccc} u_1 & u_2 & \cdots\\ u_2 & u_1 & \cdots\\ \vdots & \vdots & \ddots \end{array}\right)\quad\text{with}\quad b_{N-i}=b_{1+i}\,. \end{align} Given two matrices $A$ and $B$ in the Lie algebra $\mathrm{gl}(2N,\mathbb{R})$, their inner product is defined by \begin{align} g(A,B)=\mathrm{tr}(\left(\begin{array}{ccccc} a_{11} & \alpha\,a_{12} & \cdots & \alpha^2\,a_{1(N-1)}& \alpha\,a_{1N}\\ \alpha\,a_{21} & a_{22} & \cdots & \alpha^3\,a_{2(N-1)}& \alpha^2\,a_{1N}\\ \vdots & \vdots & \ddots & \vdots & \vdots \end{array}\right)\left(\begin{array}{ccc} b_{11} & \alpha\,b_{12} & \cdots & \alpha^2\,b_{1(N-1)}& \alpha\,b_{1N}\\ \alpha\,b_{21} & b_{22} & \cdots & \alpha^3\,b_{2(N-1)}& \alpha^2\,b_{1N}\\ \vdots & \vdots & \ddots & \vdots & \vdots \end{array}\right)) \end{align} where $\alpha$ is a positive real number. For other tangent vectors, we require that the metric is right-invariant.

The metric has the same invariance as circulant & symmetric matrices. I want to argue that the geodesic itself $U(t)$ is itself circulant & symmetric for every $t$. Is there a simple way to prove this?

After Robert's very helpful answer, I kept thinking about the following: My $U$ is actually defined by the condition $UU^\intercal=G$ where $G$ is a positive definite, circulant, symmetric matrix. Intuitively, I just thought that $U$ should be itself circulant and symmetry and therefore $U=\sqrt{G}$. However, in general there is full class of matrices $U$ satifying $UU^\intercal=G$, namely $U\to U\cdot O$ where $O\in\mathrm{SO}(N)$ with $UO(UO)^\intercal=UOO^\intercal U^\intercal=UU^\intercal=G$. In this case, there is a geodesic from $1\!\!1$ to $U$ for every $U$ in this set. I still want to argue that the $U$ from above is the one with the shortest distance and I could verify this explicitly for the case $N=2$. Is there a simple argument that generalizes this to larger $N$?

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    $\begingroup$ I don't understand the formula defining your inner product. What is $\alpha$? And why the notation GL(2N,R) when your matrices are N by N? $\endgroup$ – Yemon Choi Jun 1 '17 at 19:06
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    $\begingroup$ Also: I may have misunderstood your question, but conjugation with an appropriate complex unitary matrix simultaneously diagonalizes every circulant. Does this conjugation preserve your metric? and if so, doesn't this reduce your problem to one of finding a geodesic on a torus? $\endgroup$ – Yemon Choi Jun 1 '17 at 19:08
  • $\begingroup$ @YemonChoi: Ok, I changed it to GL(N,R) to make it consistent. $\alpha$ is a positive real number. Yes, the circulant matrices can be diagonalized. The metric is not invariant under this transformation - maybe you are right that the geodesics are just on a torus, but I'm not sure how to prove this. $\endgroup$ – LFH Jun 1 '17 at 21:05
  • $\begingroup$ @YemonChoi: The metric is compatible in the sense that it is invariant if the components of the matrix $A$ and $B$ are mapped $A_{ij}\to A_{ji}$ and $A_{ij}\to A_{(i+n)(j+n)}$. In this sense, the matrix is invariant under reflections and discrete translations (just like circulant/symmetric matrices). $\endgroup$ – LFH Jun 1 '17 at 21:11
  • $\begingroup$ @LFH: You must also be assuming that $U$ be positive definite, otherwise, there is no smooth curve joining $U$ to the identity that lies in the submanifold $C_N\subset \mathrm{GL}(N,\mathbb{R})$ consisting of circulant and symmetric $N$-by-$N$ matrices, even when the determinant of $U$ is positive. (Just look at the case $N=3$.) $\endgroup$ – Robert Bryant Jun 6 '17 at 6:54
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The answer is as follows: When $U$ is a positive-definite, symmetric, circulant matrix in $\mathrm{GL}(n,\mathbb{R})$, then there is a symmetric circulant matrix $u$ such that $U = e^u$ and the curve $\gamma(t) = e^{tu}$ (which is a positive-definite, symmetric, circulant matrix for all $t$) is a geodesic in the metric $g_\alpha$ on $\mathrm{GL}(n,\mathbb{R})$ described by the OP for each $\alpha>0$.

Here is a sketch of a proof.

First, fix $\alpha>0$ and define a vector space isomorphism $\phi:{\frak{gl}}(N,\mathbb{R})\to {\frak{gl}}(N,\mathbb{R})$ by $\phi(A_{ij}) = A'_{ij}$ where $A'_{ij} = \alpha^{d(i,j)}A_{ij}$ and where $d(i,j)=d(j,i)\ge0$ is the mininum distance from $i-j$ to an integer multiple of $N$. Let $g_\alpha$ be the right-invariant (pseudo-)Riemannian metric on $\mathrm{GL}(N,\mathbb{R})$ for which the inner product at the identity $I_N\in\mathrm{GL}(N,\mathbb{R})$ is $$ g_\alpha(A,B) = \mathrm{tr}\bigl(\phi(A)\phi(B)\bigr) = \mathrm{tr}\bigl(\phi^2(A)B\bigr)= \mathrm{tr}\bigl(A\phi^2(B)\bigr). $$ (Note that $g_\alpha$ is both left- and right- invariant if and only if $\alpha = 1$.)

Let $C_N$ denote the (vector) space of symmetric, circulant $N$-by-$N$ matrices. Then $C_N$ is closed under multiplication and forms a commutative sub-ring of the matrix ring $M_N(\mathbb{R})$ and an abelian subagebra of the Lie algebra ${\frak{gl}}(N,\mathbb{R})$. Let $C_N^+\subset C_N$ denote the set of positive definite elements of $C_N$, so that $C_N^+$ is the connected component of $C_N\cap \mathrm{GL}(N,\mathbb{R})$ that contains $I_n$. The usual matrix exponential mapping $\mathrm{exp}(u) = e^u$ induces a bijective diffeomorphism $\mathrm{exp}:C_N\to C_N^+$. Also, note that $\phi(C_N) = C_N$.

Meanwhile, calculation shows that, $\gamma:\mathbb{R}\to \mathrm{GL}(N,\mathbb{R})$ is a geodesic in the metric $g_\alpha$ if and only if $v(t) = \gamma'(t)\gamma(t)^{-1}$ satisfies the Euler equation for $g_\alpha$, namely, $$ v'(t) = - \phi^{-2}\bigl(\bigl[v(t),\phi^2(v(t))\bigr]\bigr). $$

Now, consider the curve $\gamma(t) = e^{tu}$ for $u\in C_n$. It satisfies $v(t) = \gamma'(t)\gamma(t)^{-1} = u$, and, since both $u$ and $\phi^2(u)$ belong to $C_N$, it follows that $$ 0 = v'(t) = - \phi^{-2}\bigl(\bigl[u,\phi^2(u)\bigr]\bigr) = -\phi^{-2}(0) = 0. $$ Thus, $\gamma:\mathbb{R}\to \mathrm{GL}(N,\mathbb{R})$ is a geodesic for all $u\in C_N$, as claimed.

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  • $\begingroup$ Your answer was extremely helpful and very clear. Thank you very much! In your answer, you probably meant $A'_{ij}=\alpha^{|i-j|}$ where $|i-j|$ is the shortest distance with the $i$ thought to be arranged in a circle, meaning $1\equiv N+1$, but that's just for consistency with my question. I completely agree with your proof. $\endgroup$ – LFH Jun 8 '17 at 6:39
  • $\begingroup$ @LFH: Yes, you are right that I got carried away, oversimplifying the definition of $\phi$. I'll fix that. $\endgroup$ – Robert Bryant Jun 8 '17 at 8:00

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