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The set $V= \{ A\in M_n(\mathbb{C})\ |\ A\bar A = I\}$ is a smooth submanifold of $M_n(\mathbb{C})$ with real dimension $n^2$. Proof: Consider the involution $\iota:\mathrm{GL}(n,\mathbb{C})\to \mat …

If the characteristic of the ground field is not 2, then a bilinear form B on a vector space V can be written uniquely in the form B = A + S, where A is an antisymmetric bilinear form and S is a symme …

Here is an argument showing that the answer is 'yes'. I'll let you check the details and that this result generalizes to all higher degrees. Consider the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n) …

The set $\pi^{-1}(\mathcal{O}^\mu)$ is always a smooth submanifold of $\mathcal{O}_\lambda$ (though it may well be empty). When it is not empty, it is a single orbit of $\mathrm{U}(n)\subset \mathrm{ …

If you let $I$ denote multiplication by $\sqrt{-1}$, then the two operators $I$ and $M$ on your vector space (say, $V$) satisfy $$ I^2 = -1,\qquad M^2 = 1,\qquad\text{and}\qquad IM=-MI. $$ (The former …

Here is a revised partial answer and some comments: It seems that you are asking for some kind of isomorphism between $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ that would 'have the greate …

Unfortunately, the answer is 'no'. You are basically asking whether you can simultaneously diagonalize two quadratic forms in three variables, and the answer is that, 'generically' you can (and you a …

A good class of examples of this is given by Clifford algebras: If $V$ is a real vector space with endowed with a quadratic form $q:V\to\mathbb{R}$, the algebra $Cl(q)$ is the algebra generated by th …

Yes, one can always do this. In fact, one can assume that $\det(U)=1$, as this follows by a homotopy argument, using the fact that $\pi_3(S^2)\simeq \mathbb{Z}$. Here are the details: Assume that $ …

I assume that the problem is to try to determine which pairs $(P_1,P_2)$ of positive definite Hermitian symmetric $N$-by-$N$ matrices can be written in the above form for some pair $(H_1,H_2)$ of posi …

Isn't the answer just $$ D(n;d) = n{{n+d-1}\choose{d}}- {{n+d}\choose{d+1}}= d{{n+d-1}\choose{d+1}}\quad ? $$

The answer is 'yes', because this space is the homogeneous space $\mathrm{GL}^+(2n,\mathbb{R})/\mathrm{Sp}(n,\mathbb{R})$, which is connected. Here is more detail: Let $J_n$ be the $2n$-by-$2n$ matr …

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 yo …

The difficulty is that the space $S_{2n}(\mathbb{R})$ of $2n$-by-$2n$ symmetric matrices with real entries is not irreducible under the action of $\mathrm{U}(n)$. In fact, there is a $\mathrm{U}(n)$- …

At least for $3$-by-$3$ matrices, the test for complete positivity of a matrix $A$ is not hard. Basically, you need that $A$ be positive-semi-definite and that the off-diagonal entries be non-negativ …