5
$\begingroup$

The Weyl group $W$ of $G_2$, is a group of order $12$ which is generated by the subgroup of permutations of $e_1$, $e_2$ and $e_3$ and by by the element $\tau$ which maps $(e_1,e_2,e_3)$ to $(-e_1,-e_2,-e_3)$.

We now define an action of $W$ on the flag manifold $U(4)/T^4$. A permutation of $e_1$, $e_2$ and $e_3$ acts on $g \in U(4)$ by permuting the corresponding first $3$ columns $c_1$, $c_2$ and $c_3$ of $g$. For example, $(123)$ places $c_1$ in the second column, places $c_2$ in the third column and places $c_3$ in the first column. This action descends to an action of the permutation subgroup of $W$ on the flag manifold $U(4)/T^4$.

Now choose a quaternionic structure $j$ on $\mathbb{C}^4$ which is compatible with the standard hermitian inner product $(-,-)$ on $\mathbb{C}^4$. What this means is that

$(jv,jw) = \overline{(v,w)}$ for any $v$, $w$ in $\mathbb{C}^4$.

Note that $j$ acts on $g \in U(4)$ by acting on each of the $4$ columns of $g$ simultaneously. This action descends to an action, say $\tilde{j}$, on $U(4)/T^4$, satisfying $\tilde{j}^2 = \operatorname{Id}$. Let the element $\tau$ of $W$ act on $U(4)/T^4$ by $\tilde{j}$.

One can verify that we have defined an action of $W$ on $U(4)/T^4$. Moreover, since $W$ is the Weyl group of $G_2$, it acts naturally on the corresponding flag manifold $G_2/T^2$.

My question can now be formulated. Does there exist a smooth $W$-equivariant map $f: U(4)/T^4 \to G_2/T^2$? I am hoping there is, though I have a feeling there isn't. Your help is kindly appreciated!

Edit 1: I just realized that both manifolds are $12$-dimensional. Are they diffeomorphic by any chance? Answer: they are not, see the comment by Michael Albanese below.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Unless I am mistaken, it follows from the long exact sequence in homotopy that $\pi_2(SU(4)/S(T^4)) \cong \pi_1(S(T^4)) \cong \mathbb{Z}^3$ while $\pi_2(G_2/T^2) \cong \pi_1(T^2) \cong \mathbb{Z}^2$. In particular, the manifolds are not diffeomorphic. $\endgroup$
    – Michael Albanese
    Commented Jun 24, 2020 at 13:30
  • $\begingroup$ @MichaelAlbanese, yes your argument is correct. I did not know for instance that $\pi_2(G)$ is trivial for any Lie group (is the statement correct, or does it require some assumptions?). Indeed, here is a relevant discussion <mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups>. $\endgroup$
    – Malkoun
    Commented Jun 24, 2020 at 14:54
  • $\begingroup$ It is true for all Lie groups. $\endgroup$
    – Michael Albanese
    Commented Jun 24, 2020 at 14:59
  • $\begingroup$ Your action on $SU(4)$ is not an action on $SU(4)$. For example, the permutation $(12)$ changes the determinant of an element of $SU(4)$ to $-1$. Are you also, say, multiplying the last column by $(-1)^{|\sigma|}$ for a permutation $\sigma$? $\endgroup$
    – Jason DeVito - on hiatus
    Commented Jun 24, 2020 at 22:58
  • $\begingroup$ @JasonDeVito, I fixed it now. One way to do things is indeed as you suggested. I wrote it in an equivalent way, by defining the action of the permutation subgroup on an element of $U(4)$ rather than $SU(4)$. Thank you. $\endgroup$
    – Malkoun
    Commented Jun 25, 2020 at 7:04

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .