# The curvature of the space of commuting hermitian matrices

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This is a question that I asked in the mathematics section, but I believe it may get more attention here. I am working on a project dedicated to the quantisation of commuting matrix models. In the appropriate formalism this problem is reduced to a quantisation in a curved space -- the space of commuting matrices. The general prescription for quantisation in curved space involves ambiguity of the Hamiltonian operator proportional to the scalar curvature of the curved space - hence my question.

A set of $p$ commuting $n\times n$ hermitian matrices $X^{\mu}$ for $\mu=1,\dots p$, is parametrised in terms of a set of $p$ diagonal matrices $\Lambda^{\mu}$ and an unitary matrix $U$ via:

$X^{\mu}=U\,\Lambda^{\mu}\,U^{\dagger}~~$ for $~\mu=1\dots p$,

clearly not all degrees of U contribute to this parametrisation, for example a reparametrisation $U'= D\,U$, where $D$ is a diagonal unitary matrix would result in the same set of commuting matrices. In other words only the elements of the quotient space $U(n)\,/\,U(1)^n$, which is the maximal flag manifold $F_n$, contribute to the parametrisation. The metric on the resulting curved manifold can be calculated as a pull-back of the metric on the space of hermitian matrices defined as:

$ds_{X}^2=Tr\,\left( dX^{\mu}dX^{\mu}\right)$ ,

Using that $U^{\dagger}d X^{\mu} U=d\Lambda^{\mu}+[\theta,\Lambda^{\mu}]~~$, where $\theta$ is the Maurer-Cartan form $\theta=U^{\dagger}dU$, one can write the induced metric as:

$ds^2=\sum\limits_{i=1}^n(d\vec\lambda_i)^2+2\sum\limits_{i<j}(\vec\lambda_i-\vec\lambda_j)^2\theta_{ij}\bar{\theta}_{ij}~~$, where $~~\vec \lambda_i =(\Lambda^1_{ii},\dots,\Lambda^p_{ii})$ .

Now I need the Riemann curvature of the above metric. It seems that it is convenient to work in tetrad formalism, using tetrads $E_{ij}=|\vec\lambda_i-\vec\lambda_j|\,\theta_{ij}$, for $i<j$. The problem is that $d E_{ij}$ will now contain a term proportional to $(\theta_{ii}-\theta_{jj})\wedge\theta_{ij}$ and since $\theta_{ii}$ are not part of the basis the spin curvature cannot be written easily without using the explicit parametrisation of $U(n)$. Intuitively, I know that the scalar curvature should depend only on the lambdas ($\vec\lambda_i$), and I have verified that explicitly for $SU(2)$ and $SU(3)$, however a general result seems to require some invariant way to express the pullback of the term $(\theta_{ii}-\theta_{jj})\wedge\theta_{ij}$ on the submanifold spanned by the off diagonal $\theta$'s.

I was wondering if mathematicians have explored the manifold of commuting hermitian matrices. In fact even a reference to a convenient parametrisation of the maximal flag manifold $F_n$ would greatly help me in deriving a general expression for the scalar curvature. Any comments/suggestions are welcomed.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user vesofilev
Crossposted from math.stackexchange.com/q/773476/11127

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user Qmechanic
The space of $p$ commuting $n\times n$ matrices is not connected, since a unitary transformation does not change the rank of a matrix. Each connected component corresponds to a specific choice of the ranks of the commuting matrices. The connected components are generalized flag manifolds. Since they are homogeneous spaces, they are of constant curvature and there are many ways to calculate their curvatures.The following lecture notes may be of help to you uregina.ca/~mareal/flag-coh.pdf

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user David Bar Moshe
Thanks for the comment. I have read parts of this paper before. I am a bit confused by your statement though. One can easily work out explicitly the 2×2 case: Upon redefining $\vec\lambda_1−\vec\lambda_2=\sqrt{2}\vec r$ and $\vec\lambda_1+\vec\lambda_2=\sqrt{2}\vec y$. And further change to polar coordinates in the space spanned by r⃗ one obtains:

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user vesofilev
$ds^2=d{\vec y}^2+du^2+u^2\left(d\Omega_{p-1}^2+d\Omega_2^2\right)$, where the vector arrow represents induces running from 1 to $p$ for the number of commuting matrices and $u=\vec r^2$. Now I can see that this space is not simply connected, but it is connected and it does not have a constant curvature, rather it is proportional to $1/u^2$.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user vesofilev
@vesofilev: The $SU(2)$ case is a special case. Take for example the space of one commuting matrix in $SU(3)$, the spaces obtained from the adjoint action on $diag(0,0,1)$ and $diag(0,1,1)$ are disconnected.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user David Bar Moshe
For one commuting matrix, it is just a flat space, because it is just a change of co-ordinates. The thing is you have to vary the elements of the diagonal matrix to. If we follow the logic from your last example, even $R^3$ would be disconnected. You can take two planes extended along $x$ and $y$, but separated in $z$. These planes are indeed disconnected, but the whole space isn't. I think your example is similar.

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user vesofilev
In other words the matrices $diag(0,0,1)$ and $diag(0,1,1)$ considered as points in the space of diagonal hermitian matrices (which is the same as the space of real diagonal matrices) can be connected, because the space of diagonal real matrices (which is $R^3$) is connected

This post imported from StackExchange Physics at 2014-05-04 11:28 (UCT), posted by SE-user vesofilev

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