# The normalizer of SU(n) in U(m)?

$$\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\Spin{Spin}$$Consider the special unitary group $$\SU(5)$$ and the unitary group $$\U(16)$$.

Below I specify a specfic way to embed $$\SU(5) \subset \U(16)$$:

1. First we can embed the spin group $$\Spin(10)\subset \U(16)$$. Here we choose the $$\mathbf{16}$$-dimensional spinor representation of $$\Spin(10)$$ to be also the $$\mathbf{16}$$-dimensional fundamental representation of $$\U(16)$$. Thus, the data for the representation: $$\text{\mathbf{16} in \Spin(10) as \mathbf{16} in \U(16)}$$ gives us an action of $$\Spin(10)$$ and $$\U(16)$$ on the complex vector space $$\mathbb{C}^{16}$$.

2. Then, we can embed $$\SU(5)$$ into $$\Spin(10)$$. See the discussion of this embedding of $$\SU(n)$$ into $$\Spin(2n)$$. Here we choose the complex-conjugated fundamental representation $$\overline{\mathbf 5}$$, the anti-symmetric representation $$\mathbf{10}$$ and the 1-dimensional representation $$\mathbf 1$$ of $$\SU(5)$$: $$\text{\overline{\mathbf 5} \oplus \mathbf{10} \oplus \mathbf 1 in \SU(5) as \mathbf{16} in \Spin(10)}.$$ Again, the data for the representation gives us an action of $$\SU(5)$$ on $$\mathbb{C}^{16}$$, which further gives an embedding of $$\SU(5)$$ into $$\U(16)$$.

## Question

Then my question is about the normalizer of this $$\SU(5)$$ inside $$\U(16)$$, which certainly depends on the embedding that I provided above. So what is the normalizer of this $$\SU(5)$$ inside $$\U(16)$$? My suspicion is that it may be $$\U(5)$$, but it could also be larger than $$\U(5)$$ with more $$\U(1)$$ factors if I was mistaken ….

$$\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}$$It's $$\U(5) \times \U(1) \times \U(1)$$.

We have a natural map from the normalizer of $$G$$ to the outer automorphism group of $$G$$. The outer automorphism group of $$G$$ is $$\mathbb Z/2$$, generated by the inverse transpose / complex conjugation.

This map is trivial, because if it were nontrivial, an element mapping to $$1 \in \mathbb Z/2$$ would give an isomorphism between this representation and the complex conjugate $$\mathbf 5 \oplus \overline{\mathbf {10}} \oplus \mathbf 1$$, which does not exist as these are not isomorphic.

Let us check that the kernel of this natural map, for any $$G \subseteq H$$, is $$G C_{H}(G)$$, where $$C_H(G)$$ is the centralizer of $$G$$ in $$H$$. To do this, note that every element $$h$$ of the kernel acts by conjugation on $$G$$ as an inner automorphism. By the definition of inner automorphism, there exists $$g \in G$$ such that the action of $$h$$ by conjugation on $$G$$ equals the action of $$g$$ by conjugation on $$G$$. It follows that $$g^{-1} h$$ acts trivially by conjugation on $$G$$. By definition, this means $$g^{-1} h$$ lies in the centralizer of $$G$$. So $$g = g \cdot (g^{-1}h) \in G C_H(G)$$. Conversely, every element of $$G C_H(G)$$ normalizes $$G$$ and acts by inner automorphisms on $$G$$, since $$C_H(G)$$ acts trivially and $$G$$ acts by inner automorphisms.

The centralizer is $$\U(1) \times \U(1) \times \U(1)$$ since there are three non-isomorphic irreducible representations of $$G$$, so the centralizer of $$G$$ consists of matrices acting by scalars on these three representations.

So the normalizer is the product, inside $$\U(16)$$, of this $$\SU(5)$$ and the $$\U(1) \times \U(1) \times \U(1)$$ scalars. We can check that this product is isomorphic to $$\U(5) \times \U(1) \times \U(1)$$ by letting $$\U(5)$$ act on $$\overline{\mathbf 5}$$ by the dual of the standard representation, $$\mathbf{10}$$ by the $$\wedge^2$$ of the standard representation, and $$\mathbf 1$$ trivially, and letting the two factors of $$\U(1)$$ act by multiplication on the last two representations.

• What is "this representation" in "an isomorphism betwee this representation and the complex conjugate"? Commented Feb 19, 2021 at 2:19
• @LSpice I mean $\overline{\bf 5} \oplus {\bf 10} \oplus {\bf 1}$. Commented Feb 19, 2021 at 2:22
• @wonderich Do you disbelieve the argument in my previous comment? Commented Feb 19, 2021 at 14:00
• @wonderich Yes, because the centralizer consists of matrices that act as scalars on each irreducible representation (Schur's lemma), and thus is $U(1) \times U(1) \times U(1)$ (and because of the outer automorphism argument). Commented Feb 21, 2021 at 20:11
• @wonderich Sorry, by $Z(G)$ I meant the centralizer, and by $G Z(G)$ the set of all products of an element of $G$ with an element of the centralizer of $G$. Commented Feb 23, 2021 at 19:17