standard form of antisymmetric matrix

Suppose $J$ is standard symplectic matrix =$\begin{bmatrix} 0 & I_d\\ -I_d & 0\end{bmatrix}$. Now $\theta=(\theta_{i,j})$ be any invertible $d\times d$ skew symmetric matrix. Then there exists an invertible matrix $T_\theta$ such that $T_\theta^tJT_\theta = \theta$. This $T_\theta$ is of course not unique. Does anyone know any explicit formula for one of $T_\theta$s consisting $\theta_{i,j}$'s?

• Note that you want $\theta$ to be an invertible $2d \times 2d$ matrix, not $d \times d$. – Robert Bryant Aug 14 '16 at 18:57

This is not really an answer, but a caution that there probably is no answer in the form that you want it.

It's not completely clear what criteria you want 'an explicit formula' to satisfy, but if you want continuity, then you are mostly out of luck.

Let $A_d$ denote the set of invertible $2d$-by-$2d$ skew-symmetric matrices, considered as an open subset of the vector space of all $2d$-by-$2d$ skew-symmetric matrices and let $\mathrm{GL}(2d,\mathbb{R})$ denote the set of invertible $2d$-by-$2d$ matrices, considered as an open subset of the vector space of all $2d$-by-$2d$ matrices.

We have a smooth, surjective, submersion $\pi: \mathrm{GL}(2d,\mathbb{R})\to A_d$ given by $$\pi(a) = a^t\, J\, a.$$ Note that this mapping $\pi$ makes $\mathrm{GL}(2d,\mathbb{R})$ into a smooth, principal, right $\mathrm{Sp}(d,\mathbb{R})$-bundle over $A_d$.

It appears that what you are asking for is an 'explicit' map $T:A_d\to \mathrm{GL}(2d,\mathbb{R})$ that satisfies $\pi\bigl(T(\theta)\bigr) = \theta$ for all $\theta\in A_d$.

It's easy to write down such a map $T$ when $d=1$ (exercise), and, in fact, we can arrange that $T$ is even smooth in this case.

However, it's not hard to show that, for $d>1$, there cannot be any continuous (let alone smooth) mapping $T:A_d\to \mathrm{GL}(2d,\mathbb{R})$ that does the job. The reason is that the bundle $\pi: \mathrm{GL}(2d,\mathbb{R})\to A_d$ is known not to be trivial when $d>1$. (One can prove this using characteristic classes.)

Thus, any 'explicit' map $T$ that you write down (when $d>1$) to solve your problem cannot be continuous. In particular, to construct such a mapping, you will need to choose some kind of locus of discontinuity, and it will not be topologically trivial. I imagine that you would want your 'explicit' formula for $T_\theta$ to be algebraic in the entries of $\theta$, say, in which case, the algebraic formula will have to have a fairly complicated locus of discontinuity, so it's not likely to be a very nice formula.

Finally, let me point out that you don't generally expect these matrix problems to have simple explicit solutions. For example, consider the 'simpler' problem of finding an 'explicit' formula for the (unique) positive definite square root $\sqrt Q$ of a positive definite symmetric matrix $Q$. It is known that $\sqrt Q$ is a smooth function of $Q$ in this case, but, when the dimension $d$ is greater than $2$, there is no easy way to write down the formula explicitly. (The case $d=2$ is exceptional, and it can be done then.)

• Thanks for excellent answer. Is this still possible? if i have a continuous path of skew symmetric matrices $\theta_t \in A_d$ say indexed by closed interval, can I choose a continuous path $a_t \in GL(2d, \mathbb{R})$ such that $T(\theta_t ) = a_t$ for all $t$? – zionnn Aug 15 '16 at 18:10
• @Zionnn: Well, for each continuous path $\theta_t$ in $A_d$ there certainly exists a continuous path $a_t$ satisfying $\pi(a_t) = \theta_t$. Whether you can choose $a_t$ depends on what you mean by 'choose'. For any noncontinuous map $T:A_d\to \mathrm{GL}(2d, \mathbb{R})$ that satisfies $\pi\circ T = \mathrm{id}$, there will exist a continuous path $\theta_t$ in $A_d$ such that $a_t = T(\theta_t)$ is not continuous. – Robert Bryant Aug 16 '16 at 1:19