Consider the space of $n\times n$ complex matrices, and equip it with its Lebesgue measure $dX$, seen as a $2n^2$-dimensional real vector space [edit: or better, a complex vector space (see the answer of Leonid Kovalev below)]. I was wondering if their is an explicit formula for the restriction of the form $dX$ to the submanifold of normal matrices, that is the matrices which satisfy $X^*X=XX^*$, where $*$ stands for the transconjugation.

Thanks in advance.

EDIT : After the comment of GEdgar : Is there some way to put an associated Riemaniann metric to the Lebesgue measure then ? And to compute it explicitly ?

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    $\begingroup$ That manifold has measure zero, so you need to say something else. $\endgroup$
    – GEdgar
    Dec 14 '11 at 16:31
  • $\begingroup$ Oups, you're right. So... no "natural" measure to put on that manifold ? $\endgroup$
    – Student
    Dec 14 '11 at 17:12

An old question, but it's worth recording an answer: the set of all normal matrices, denoted $N$ below, is not a submanifold of $\mathbb R^{2n^2}$.

Clearly, $N$ contains $0$ and is invariant under scaling: $N=\lambda N$ for all $\lambda > 0$. We need a simple

Lemma. Any scaling-invariant submanifold $M\subset \mathbb R^m$ that contains $0$ is a linear subspace.

Proof. Indeed, let $P$ be the tangent space $T_0 M$, considered as a subspace of $\mathbb R^m$. If $x\in M\setminus P$, then the segment containing $0$ to $x$ lies in $M$ and forms a positive angle with $P$, a contradiction. Thus $M\subset P$. Since $M$ has the same dimension as $P$, it is open in the topology of $P$. Considering that $0\in M$ and $M$ is scale-invariant, we conclude that $M=P$. $\square $

It remains to note that $N$ is not a linear subspace unless $n=1$. A $2\times 2$ example suffices for this: $\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$ and $\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$ are normal, but $\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ is not.

  • $\begingroup$ Very interesting, thanks ! $\endgroup$
    – Student
    Aug 11 '12 at 2:49
  • $\begingroup$ My question is doomed anyway. I've find my answer, to consider the Riemannian metric on it as a complex manifold is the answer I was looking at... $\endgroup$
    – Student
    Aug 11 '12 at 2:50

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