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Given $G \in SU(4)$, what are the level sets of the function $F:SU(n)\rightarrow \mathbb{R}$ defined by $F(V) = |tr(G^{\dagger}V)|^2$?

Can they be written only in terms of abstract linear maps, not in terms of the components of $V$ and $G$ in some basis?

Cross-posted after no answer on MSE.

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  • $\begingroup$ What does the dagger mean? Also, where does $V$ live? $\endgroup$ Sep 9, 2015 at 14:38
  • $\begingroup$ The dagger is the conjugate transpose $\endgroup$
    – Benjamin
    Sep 9, 2015 at 16:10

2 Answers 2

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First of all, the $G$ is a red herring. Multiplying on the left by $G$ (which is an isometry of $\mathrm{SU}(4)$ with respect to any left-invariant metric), we have $F(GV) = \bigl|\mathrm{tr}(V)\bigr|^2$, so it suffices to understand what is going on when $G=I$. So, from now on, set $G=I$.

Second, the function $F(V) = \bigl|\mathrm{tr}(V)\bigr|^2$ is, of course, constant on conjugacy classes, and we have $$ F(V) = \bigl|\lambda_1+\lambda_2+\lambda_3+\lambda_4\bigr|^2 $$ where $\lambda_i\in S^1$ are the eigenvalues of $V$, which satisfy $\lambda_1\lambda_2\lambda_3\lambda_4=1$.

Thus, $F$ reaches the maximum value of $16$ when all of the eigenvalues are equal, which happens only at the $4$ elements in the center of $\mathrm{SU}(4)$ (these are nondegenerate maxima), and it reaches the minimum value of $0$ (also a critical value) only when the eigenvalues of $V$ are $\lambda, \bar\lambda, -\lambda, -\bar\lambda$, where $|\lambda|=1$. Thus, the zero level set is a union of a $1$-parameter family of conjugacy classes, most of which have dimension $12 = 15-3$, but there are two conjugacy classes (when $\lambda = 1$ and $\lambda = i$) that have dimension $8$ (and are copies of the Grassmannian of complex $2$-planes in $\mathbb{C}^4$). Still, the zero level set is smooth because it's also the set of matrices in $\mathrm{SU}(4)$ with trace $0$, and this is a smooth submanifold of dimension $13$.

There is only one other critical value of $F$, which is $4$, and the critical points of $F$ that have this value are the matrices $V$ with eigenvalues $\lambda,\lambda,\lambda,-\lambda$ where $\lambda^4 =-1$. These constitute $4$ conjugacy classes, each of which is a copy of $\mathbb{CP}^3$ embedded in $\mathrm{SU}(4)$, but, of course, the entire level set $F(V) = 2$ is a $2$-parameter family of conjugacy classes, all of which have dimension $12$ except the $4$ that I listed above.

In particular, note that, for $4<c<16$, the level set $F(V) = c$ is a disjoint union of $4$ spheres of dimension $14$, while, for $0<c<4$, the level set $F(V)=c$ is a (trivial) circle bundle over the level set $F(V)=0$.

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  • $\begingroup$ Again, great and thanks. The level sets of this function don't form a foliation of $SU(4)$ then I'm guessing. $\endgroup$
    – Benjamin
    Sep 10, 2015 at 13:24
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    $\begingroup$ Well, the level sets of $F$ are smooth hypersurfaces except for the levels $0$ (smooth and connected, of dimension $13$), $4$ (singular at some places but connected and of dimension $14$ at smooth points), and $16$ (consisting of $4$ points). All the other level sets form a smooth foliation of the remainder of $\mathrm{SU}(4)$. $\endgroup$ Sep 10, 2015 at 13:31
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If $X^\dagger$ denotes transposed conjugate, since $tr(X^\dagger Y)$ is the Hermitian inner product on $\mathfrak{gl}(4,\mathbb C)$, the level sets are codimesion 1 circular cylinders (circles in the complex plane $\mathbb C.G$ times the orthogonal complement of $G$) intersected with $SU(4)$.

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