Timeline for Largest subgroup of $SU(n)$ for which the adjoint action preserves specific inner product on $\mathfrak{su}(N)$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2015 at 17:26 | vote | accept | Benjamin | ||
| Apr 22, 2015 at 17:12 | answer | added | Robert Bryant | timeline score: 2 | |
| Apr 22, 2015 at 15:23 | comment | added | Benjamin | I mean that we are assuming it if the $SU(2)$ example is to represent the general case. The fact that $P,Q$ are maximal in this case is clear as you point out. In the $SU(4)$ case it is far less clear to me that they will be maximal but I expect that's because I lack the intuition that lead you to think they are in the first place! | |
| Apr 22, 2015 at 9:21 | comment | added | Robert Bryant | I couldn't figure out how to login in the chat discussion, so I'm continuing here: I didn't assume that $P$ and $Q$ are maximal, I wrote that I thought it was likely. In the particular case $n=2$ above, $\mathrm{SU}(2)$, which is $3$-dimensional, doesn't have any $2$-dimensional subgroups, so $P$ and $Q$ (which are $1$-dimensional) have maximal dimension by default. Is that what you are asking about? | |
| Apr 22, 2015 at 3:20 | comment | added | Benjamin | I understand your example. However, you are assuming that $P$ and $Q$ are maximal in the sense you described above. It is this step I'm failing to complete the details of. Why should we expect only the stabiliser subgroups of $A,B$ to leave the inner product unchanged? | |
| Apr 22, 2015 at 1:46 | comment | added | Robert Bryant | Well, take a simple example: Let $n=2$ and let $A$ and $B$ be perpendicular unit vectors in ${\frak{su}}(2)$. Then $P$ and $Q$ (as defined in my first comment) are distinct $S^1$-subgroups and their union generates all of $\mathrm{SU}(2)$. It follows that $S_{A,B}$, which is easily seen to be a codimension-$2$ subset of $\mathrm{SU}(2)$, cannot be a subgroup (though it does contain $P$ and $Q$). Instead, it is a $1$-parameter family of cosets of an $S^1$-subgroup. | |
| Apr 21, 2015 at 23:53 | comment | added | Benjamin | How can one see that it's not a group? Do we really mean it's not a group or just not a sub Lie group. Based on this, how do we hope to assign it a dimension, can we know that it's a smooth manifold at least? | |
| Apr 21, 2015 at 19:56 | comment | added | Robert Bryant | Generally, it won't be true that the set $S_{A,B}$ consisting of all $U\in\mathrm{SU}(n)$ that satisfy $K\bigl(A,\mathrm{Ad}_U(B)\bigr)=0$ is a subgroup of $\mathrm{SU}(n)$, so the best you can hope for is to look for the 'largest' (i.e., highest dimension) subset $H\subset S_{A,B}$ that forms a subgroup, i.e., is closed under multiplication and inverse. | |
| Apr 21, 2015 at 18:29 | comment | added | Benjamin | Why do you say most likely they will be maximal? Won't the group of all such $U$ be the largest. By maximal do you mean, one we can't extend to a larger supergroup? Or rather, simply the such largest group. Perhaps I am confused. | |
| Apr 21, 2015 at 11:53 | comment | added | Robert Bryant | Why do you think that there is one largest subgroup? For example, I imagine that, for the generic pair $A,B\in{\frak{su}}(n)$, there will be two subgroups, $P = \{ U\in\mathrm{SU}(n)\ |\ \mathrm{Ad}_U(A) = A\}$ and $Q = \{ U\in\mathrm{SU}(n)\ |\ \mathrm{Ad}_U(B) = B\}$ that satisfy your condition and, most likely, they will each be maximal, even though they could be of different dimensions. | |
| Apr 21, 2015 at 1:44 | history | edited | Benjamin | CC BY-SA 3.0 |
edited body
|
| Apr 21, 2015 at 1:22 | history | edited | Benjamin | CC BY-SA 3.0 |
edited title
|
| Apr 21, 2015 at 1:16 | history | asked | Benjamin | CC BY-SA 3.0 |