Is there anything "nice" about the set of normal matrices (over $\Bbb R$ and $\Bbb C$?)

Normal matrices are of course useful to any linear algebra buff, not least because of the spectral theorem. However, taken as a whole, they tend to have some not-so-nice properties. For example:

• unlike the unitary matrices, they are not closed under multiplication. That is, $A$ and $B$ being normal does not guarantee that $AB$ is normal
• unlike the Hermitian matrices, they don't form a linear space. That is, $A$ and $B$ being normal does not guarantee that $A+B$ is normal

I'm wondering, then, if there are any nice properties of this set, or any good ways of thinking of how the set of normal matrices sits within $\Bbb C^{n\times n}$, or perhaps $\Bbb R^{n \times n}$.

Here's the way I see it: the normal matrices are the zero-set of the quadratic map $\phi:\Bbb F^{n \times n} \to \Bbb F^{n \times n}$ ($\Bbb F \in \Bbb{\{R,C\}}$) given by $$ \phi(A) = A^*A - AA^* $$ The set is closed. It is (I think) a smooth manifold in $\Bbb R^{n \times n}$, and it is connected in $\Bbb C^{n \times n}$.

Is it also connected in $\Bbb R^{n \times n}$? What is its dimension as a manifold in $\Bbb R^{n \times n}$? Do we know anything about its topological structure (Cohomology, for instance)? Is there some underlying algebraic structure to the "space" of normal matrices? Is there some interesting connection to draw between the unitary matrices and the Hermitian matrices that comes out of this analysis (the exponential map comes to mind)?

Any input here is appreciated. Thanks for reading :)

Solution 1:

Let me summarize my comments in an answer. I don't have answers to all the questions but I don't expect (myself) to really come up with any more, so here goes.

First, I am mostly thinking about this space as the algebraic variety over $\mathbb{R}$ (complex conjugation is not $\mathbb{C}$-scalar) cut out by the $n^2$ (in the $\mathbb{R}$ case) or $2n^2$ (in the $\mathbb{C}$ cae) equations given by $$A A^* - A^* A = 0.$$

This is a $\mathbb{R}$-conic variety in both cases, so in particular every point has a straight path to 0, so the space is path-connected in both cases.

Further, the only conic varieties which are smooth varieties (i.e. manifolds) are those cut out by linear equations. To see this, recall that the singular points are those where the Jacobian does not have full rank. Since the variety is cut out by homogeneous equations (i.e. it is conic) of non-linear degree, the Jacobian must vanish at zero. Further, zero is a point on the variety, so it is singular.

EDIT: Also, because it is a conic variety it is contractible, so homotopy equivalent to a point, so it has trivial topology. A more interesting question might be to ask about its projectivization, though I don't have much on that front.

Finally, one can compute the dimension by using the action of the unitary group on the normal matrices, at least in the case of $\mathbb{C}$. The special unitary group acts on the normal matrices by conjugation. Let us compute a "big" orbit, i.e. the orbit of a diagonal matrix with distinct diagonal entries. Only diagonal matrices can stabilize this point, and the diagonal matrices in the unitary group are isomorphic to $\mathbb{C}^*)^n$, i.e. have dimension $n$ , and so this orbit has dimension $\dim U(n) - n = n^2 - n$. Further, the parameter space for the orbits consists of parameterizing the diagonal entries; since the subset of this parameter space where the entries are distinct is a Zariski open locus, its dimension is equal to the parameter space, which is $2n$ (remember, real dimension). Thus the total space has dimension $n^2 + 2n - n = n^2 + n$. I mean dimension as a variety, meaning where the variety is smooth, it is locally of that dimension.

Hopefully all the above is correct.

EDIT: This MO post links to a few papers which have further answers: the normal variety is not orientable for $n \geq 2$, and is irreducible, etc.