# Why must $U(1)$ matrices commute with $SU(2) \times SU(3)$ matrices in embedding within $SU(5)$?

I'm a physicist taking a groups course.

I can believe that the direct sum of the fundmental representations for the $$SU(2)$$ and $$SU(3)$$ matrices will work as an embedding of the $$SU(2) \times SU(3)$$ subgroup within the fundamental representation of $$SU(5)$$ (since you have a 5x5 matrix composed of a 2x2 block and a 3x3 block so obviously they act on separate vector subspaces as in the direct product).

However it is often said that to embed $$SU(2) \times SU(3) \times U(1)$$ we can just also use a representation of 5x5 matrices for $$U(1)$$ which commutes with the above matrices. I don't understand how this commuting constraint means that the $$U(1)$$ part of the subgroup acts on a 'separate vector space' like the direct product suggests?

• I suggest that instead of the word 'separate' you use 'transverse' which in this case means that they have zero intersection but are both contained in a common vector space that they together span. As for the construction, think of block-diagonal matrices and a diagonal action of U(1). Can you figure out which diagonal action to use here? Commented Jan 6, 2022 at 22:37
• Did you understand the solution? Commented Jan 17, 2022 at 23:31

First of all, please, never use the terminology 'a separate vector space.' It is meaningless. What you really meant to say is that you have a direct sum decomposition $${\mathbb C}^5=V\oplus W$$, where $$V, W$$ are (complex-linear) subspaces of $${\mathbb C}^5$$ of dimensions 2 and 3 respectively. Concretely, you can take $$V=\{(z_1,...,z_5): z_3=z_4=z_5=0\}, W= \{(z_1,...,z_5): z_1=z_2=0\}.$$ These are the subspaces that you call 'separate.' Then you take standard (fundamental) representations of $$SU(2)$$ (acting on $$V$$) and of $$SU(3)$$ (acting on $$W$$) and define a faithful representation $$\rho: SU(2)\times SU(3)\to SU(5)$$ sending each pair of matrices $$(A, B)$$ (with $$A\in SU(2), B\in SU(3)$$) to the block-diagonal matrix $$\left[\begin{array}{cc} A&0\\ 0&B\end{array}\right]$$ Given all this, you are effectively asking how to extend the representation $$\rho$$ to a faithful representation $$\hat\rho: SU(2)\times SU(3)\times U(1)\to U(5).$$ This is my reading of the sentence "I don't understand how this commuting constraint means that the 𝑈(1) part of the subgroup acts on a 'separate vector space' like the direct product suggests?" (which, on its face, does not make sense even on the level of syntax). I think, your confusion stems from the invalid assumption that in order to construct a faithful representation of a direct product of groups you have to have a corresponding direct sum decomposition of the corresponding vector space.
The extension $$\hat\rho$$ is constructed by sending each matrix $$C\in U(1)$$ (which is just a single unitary complex number $$t$$) to the corresponding diagonal 5-by-5 matrix $$\hat\rho(t)= \mathrm{diag}(t,t,t,t,t).$$ For matrices $$A\in SU(2)$$ and $$B\in SU(3)$$, I use the same representation $$\rho$$ as above, of course: $$\hat\rho(A)=\rho(A), \hat\rho(B)=\rho(B).$$
It is an easy exercise to check that $$\hat\rho(t)$$ commutes with all the matrices $$\rho(A), \rho(B)$$ as above, thus, resulting in a representation $$\hat\rho: SU(2)\times SU(3)\times U(1)\to U(5).$$ With a bit more thought, you will check that $$\hat\rho$$ is faithful. The key to this is the observation that for every two matrices $$A, B$$ as above and for each $$t\in U(1)$$, $$\rho(A,B)=\hat\rho(t)$$ if and only if $$t=1$$, $$A=I_2, B=I_3$$. (Incidentally, this argument will fail if I were to try a similar construction with $$SU(2)\times SU(2)\times U(1)$$ and a 4-dimensional representation.)