I'm reading an article which I cannot understand a paragraph very well.

$T$ is a maximal torus of $SU(k+1)$ acting linearly on $\mathbb{C}^{k+1}$. And here is what is written that I cannot fully understand:

Note that the normalizer $N=N(T)$ of $T$ in $O(2k+2)$ normalizes the centralizer $C(T)$ of $T$ in $O(2k+2)$. Assume $k\geq 2$. Since $C(T)$ is a maximal torus of $U(k+1)$ which is also a maximal torus of $O(2k+2)$, we see that $C(T)$ is the identity component of the normalizer $N$. Moreover, $N$ is generated by $C(T)$ , the complex conjugation $c$ and the symmetric group $S_{k+1}$ of permutations of complex coordinates.

First of all, I cannot see why $C(T)$ should be a maximal torus of $U(k+1)$. I can see why it should be connect, since it is the union of maximal tori in $U(k+1)$ containing $T$. My supervisor also said that it is because a element in $C(T)$ lies on $N(T)$, and we can assume, using complex conjugation $c$, that every element $n\in N(T)$ is complex-linear on $\mathbb{C}^{k+1}$ (and I cannot see why). And that, up to permutation, $n$ preserves the decomposition $\mathbb{C}^{k+1}=\mathbb{C}\oplus\cdots\oplus \mathbb{C}$ (which I also cannot see why). And furhter he said that such an element must Lie on the maximal torus $S^1\times\cdots\times S^1\subset U(K+1)$; factorization compatible with the decomposition $\mathbb{C}^{k+1}=\mathbb{C}\oplus\cdots\oplus \mathbb{C}$ (and, again, I cannot see why).

Given this, why should $C(T)$ be the indentity component of $N(T)$?

  • 1
    $\begingroup$ Why don't you calculate? Take $k=2$. Take the diagonal maximal torus in ${\rm SU}(3)$; it is 2-dimensional. Compute its centralizer in $U(3)$; it is the 3-dimensional diagonal torus. Clearly this 3-dimensional torus is a maximal torus in $U(3)$ and a maximal torus in ${\rm SO}(6)$. You continue calculations.... $\endgroup$ Commented Jul 5, 2020 at 10:36
  • $\begingroup$ I have calculated it to the usual torus in SU(n), but I would like to understant it more geometrically/generally, for an arbitrary maximal torus of $SU(n)$. I know that all need can be stracted from those calculations, but I would like to understand the techinics used by the author, which are much more general. $\endgroup$
    – Gomes93
    Commented Jul 5, 2020 at 19:44
  • 3
    $\begingroup$ It is the same! You take an arbitrary maximal torus $T\subset {\rm SU}(n)$, you consider its natural complex representation in $V:={\Bbb C}^n$, this representation uniquely decomposes in a direct sum of one-dimensional complex representations, you take eigenvectors as basis vectors, and you get the "usual" diagonal torus. Then you should calculate! $\endgroup$ Commented Jul 6, 2020 at 8:32


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.