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Let $G=SU(k)\times T^1$, $S$ a subgroup of the center of $SU(k)$ ($Z(SU(k)\cong \mathbb{Z}_k$) and $\eta$ a homomorphism from $S$ into $T^1$. Suppose $(S, \eta)$ denotes the subgroup of $G$ contains elements of the form $(s , \eta(s))$ for all $s\in S$. We know that $SO(2k)$ has some subgroups in the form $G/(S,\eta)$ (for example $U(k)$). It seems (for the reasons not mentioned here) $SO(2k)\times T^1$ has no subgroup in the above form (nontrivial $S$) which is not contained in $SO(2k)$, is it true?
*$SO(2k)\cong SO(2k)\times \{e\}$

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I'm not sure I understand the question. There is always the subgroup $\{(-I, -1), (I,1)\}\cong \mathbb{Z}/2$ which is not contained in $SO(2k)$. If you're asking whether or not the quotient of $SO(2k)\times T^1$ by this subgroup is isomorphic to the quotient $SO(2n)/\pm I \times T^1$, I'd have to think about it. –  Jason DeVito Jan 25 '13 at 13:00
    
My question is about the quotients of $G=SU(k)\times T^1$ by it's central subgroups in the form $(S,\eta )$ described above, precisely, can $(SU(k)\times T^1)/(S,\eta )$ be a subgroup of $SO(2k)\times T^1$ but not contained in $SO(2k)$? $S$ is a nontrivial subgroup of $Z(SU(k)$. –  PGMath Jan 26 '13 at 7:08
    
Can $\eta$ be trivial? I think $SU(4)/\mathbb{Z}_4$ lies in $SO(15)$, but no smaller. –  Jason DeVito Jan 26 '13 at 16:00
    
Yes, $\eta$ can be trivial. I guess representation theory (checking irreducible representations of smallest dimension for $SU(m)$) is very helpful to show that there is no such subgroup, but I have some troubles in details. –  PGMath Jan 26 '13 at 20:13
    
Are you familiar with www-math.univ-poitiers.fr/~maavl/LiE/form.html? It gives a way of answering a lot of these kinds of questions. I'll post an answer in a second. –  Jason DeVito Jan 26 '13 at 22:19

1 Answer 1

Since you allow $\eta$ to be trivial, a counterexample is provided by $G = SU(4)$ with $S = Z(SU(4))$.

So, in other words, I'm claiming $SU(4)/S \times T^1$ is not isomorphic to a subgroup of $SO(8)$. In fact, I'm claiming more: $SU(4)/S$ is not isomorphic to a subgroup of $SO(8)$. The proof will use some representation theory.

Recall the Dynkin Diagram of $SU(4)$ is $A_3$, consisting of 3 simple roots arranged in a line. For a general Lie group, the irreducible representations are uniquely characterized by their highest weight vector with can be given in terms of natural numbers over each of the 3 simple roots.

Using http://www-math.univ-poitiers.fr/~maavl/LiE/form.html, computing the dimension of a module, we see that $SU(4)$ has exactly 3 irreducible representations of dimension less than or equal to $8$, given by putting $0$s over 2 of the simple roots and a 1 over the remaining simple root.

Putting a $1$ over either the first or last root gives a 4-d rep. The first of these is the standard rep while the second is the first precomposed with complex conjugation.

Said another way, the first rep is $A\cdot v = Av$ while the second is $A\cdot v = \overline{A} v$.

Putting a $1$ over the middle simple root corresponds to the representation given by the usual double cover $SU(4)\rightarrow SO(6)$.

General representation theory tells us that a sum of representations if real iff its a sum of real irreducible plus pairs of other representations and their conjugates. In particular, the only nontrivial homomorphisms from $SU(4)$ to $SO(8)$ are the standard homomorphism coming from interpreting $\mathbb{C}$ as $\mathbb{R}^2$ and the double cover $SU(4)\rightarrow SO(6)\subseteq SO(8)$.

Now, we simply need to check the kernels of both of these and make sure they are not all of $S$. But, the first homomorphism has trivial kernel and the second has kernel generated by $-I$, so neither is all of $S$. It follows that $SU(4)/S$ is not isomorphic to a subgroup of $SO(8)$.

(Incidentally, the adjoint action of $SU(4)$ on its algebra gives an embedding of $SU(4)/S$ into $SO(15)$. I'm not sure if this is the smallest $SO(n)$ it embeds into or not.)

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@DeVito: Thanks to your nice example. Its details was very helpful for me. This example encourage us, as I said at first, that such subgroup $G$ cannot exist. I need to check all possible situations (i.e. the homomorphism to be trivial or nontrivial) to get a general answer. –  PGMath Jan 27 '13 at 7:57
    
@DeVito: The irreducible complex representations of the simply connected simple groups of are listed in the table 1 in link‌​, see also the discussion link, I think these will help to get general answer. –  PGMath Jan 27 '13 at 8:08

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