# 10-dimensional and 15-dimensional matrix representations of $SU(5)$: explicit 24 Lie algebra generators

There are some previous discussions in this post Representation of the $\rm SU(5)$ model in GUT which confused me. So I want to follow up with a new question.

It is easy to write down the 5-dimensional matrix representations of $$SU(5)$$ with 24 Lie algebra rank-5 matrix generators as:

## My question

is that based on the fact of $$5 \times 5 = 10_A + 15_S$$

How do we write down the 10-dimensional and 15-dimensional matrix representations of $$SU(5)$$?

• 10-dimensional matrix representations of $$SU(5)$$ with 24 Lie algebra rank-10 matrix generators.

• 15-dimensional matrix representations of $$SU(5)$$ with 24 Lie algebra rank-15 matrix generators.

Warning: Note that the $$10_A$$ is not just the rank-5 antisymmetric matrix as Lie algebra generators because that only gives 10 such matrices which generate the $$SO(5)$$ instead of $$SU(5)$$.

• Crossposted to math.stackexchange.com/q/4214397/11127 Aug 2 '21 at 14:25
• Yes thanks- I thought math and physics people could provide different types of thinkings - which indeed people do. :) Aug 2 '21 at 14:26

1. Let's for simplicity sketch how the construction goes for the Lie group $$U(5)$$ and leave it for the reader to modify it to $$SU(5)$$.

2. OP is interested in realizing the group representations $${\bf 10}~:=~{\bf 5}\wedge{\bf 5}~=~\begin{array}{r} [~~]\cr [~~] \end{array} \qquad\text{and}\qquad {\bf 15}~:=~{\bf 5}\odot {\bf 5}~=~[~~]~[~~],\tag{1}$$ where $${\bf 5}=[~~]$$ denotes the defining/fundamental representation of $$U(5)$$. (NB: In this answer we often identify a representation with its vector space.)

3. Together they form a 25-dimensional reducible tensor representation $${\bf 25}~:=~{\bf 5}\otimes{\bf 5}~=~{\bf 5}\wedge{\bf 5}~\oplus~{\bf 5}\odot{\bf 5}. \tag{2}$$ Here $$\otimes$$ denotes the standard (un-symmetrized) tensor product.

4. Explicitly, the tensor representation $$R:~ U(5)~\to~ {\rm End}({\bf 5}\otimes{\bf 5}),\tag{3}$$ is given as $$R(g)(\sum_iv^i_L\otimes v^i_R)~=~\sum_igv^i_L\otimes gv^i_R ,\tag{4}$$ where $$g\in~ U(5), \qquad v^i_L,v^i_R~\in~{\bf 5}.\tag{5}$$

5. The corresponding Lie algebra representation $$r:~ u(5)~\to~ {\rm End}({\bf 5}\otimes{\bf 5}),\tag{6}$$ is given as $$r(x)(\sum_iv^i_L\otimes v^i_R)~=~\sum_ixv^i_L\otimes v^i_R + \sum_iv^i_L\otimes xv^i_R,\tag{7}$$ where $$x\in~ u(5), \qquad v^i_L,v^i_R~\in~{\bf 5}.\tag{8}$$ By choosing a basis for $${\bf 5}$$, it is then in principle possible to calculate a $$25\times 25$$ matrix representation of the basis elements for the Lie algebra $$u(5)$$.

6. The tensor representations (3) and (6) respect the splitting (2) into OP's sought-for representations (1). This is in principle the answer to OP's question.

7. On the other hand, OP considers the 25-dimensional Lie algebra $$u(5)=u(5)_A\oplus u(5)_S \tag{9}$$ of anti-Hermitian $$5\times 5$$ matrices, which separates into a 10-dimensional subspace $$u(5)_A$$ of real antisymmetric matrices, and a 15-dimensional subspace $$u(5)_S$$ of imaginary symmetric matrices.

8. The adjoint representation $${\rm Ad}: ~U(5)~\to~ {\rm End}(u(5)),\tag{10}$$ is given by \begin{align} {\rm Ad}(g)x~:=~&gxg^{-1}, \cr g~\in~&U(5), \qquad x~\in~u(5),\end{align}\tag{11} acts on the Lie algebra $$u(5)$$, but it does not respect the splitting (9).

• thanks so much for this +1! Aug 1 '21 at 19:03
• But sorry in which equations do you obtain the 10-dimensional matrix representations of 𝑆𝑈(5) and the 15-dimensional matrix representations of 𝑆𝑈(5) ? Aug 1 '21 at 19:04
• This part is nice "𝑢(5)=𝑢(5)𝐴⊕𝑢(5)𝑆(9) of anti-Hermitian 5×5 matrices, which separates into a 10-dimensional subspace 𝑢(5)𝐴 of real antisymmetric matrices, and a 15-dimensional subspace 𝑢(5)𝑆 of imaginary symmetric matrices." but we hope to make these rank-10 and rank-15 matrices explicitly Aug 1 '21 at 22:30
• I updated the answer. Aug 2 '21 at 14:55
• Thanks so much for the great answer --- could you clarify: " adjoint representation acts on the Lie algebra 𝑢(5), but it does not respect the splitting $𝑢(5)=𝑢(5)_𝐴⊕𝑢(5)_𝑆$." What do we learn from the properties here? (I admire you including the discussions, points by points, so I could easily follow.) Aug 26 '21 at 15:43

Unfortunately, despite my near-promise in the question you quote, I don't know of a source that computes these bulky sets of 24 10×10 and 15×15 extraordinarily sparse matrices. The best I could do is illustrate for you the compact answer of @Qmechanic 's (2) and make sure you visualize it the way I do (and everyone should, arguably).

I will use your $$\lambda_1$$ as an example of the 5×15 generator in your non-standard normalization, $$A_1$$ for the corresponding 10×10 one, and $$S_1$$ for the 15×15 one. But, alas!, I won't even get to computing those, but just the reducible 25×25 coproduct one, $$A_1\oplus S_1$$, $$\Delta (\lambda_1)_{25}= \lambda_1\otimes 1\!\!1 _5 + 1\!\!1 _5 \otimes \lambda_1= A_1\oplus S_1 .$$

My convention for tensor products is "right-into-left", that is, the right tensor factor vectors/matrices multiplies the left vector/matrix numerical entries.

The above coproduct then is a straightforward block matrix, where I write the 5×5 blocks compactly, symbolically, $$\Delta (\lambda_1)_{25}= \begin{bmatrix} \lambda_1 & 1\!\!1 _5 &0&0&0 \\ 1\!\!1 _5 & \lambda_1 &0&0&0\\ 0&0&\lambda_1 &0&0 \\ 0&0&0&\lambda_1&0 \\ 0&0&0&0&\lambda_1 \end{bmatrix}.$$

As illustrated in both answers to the $${\mathfrak su} (2)$$ example of your choice, an orthogonal similarity Clebsch transformation effects a basis change from this uncoupled to the coupled basis, $$\begin{bmatrix} A_1&0\\0&S_1\end{bmatrix},$$ and likewise for all 23 remaining generator 25×25 matrices like this one. I would not dream of producing this Clebsch matrix, since, as I said in my answer you cite, it's a project.

How does this coproduct matrix act on a simple (too simple!) sample vector? Let's write the column vectors as transposes of row vectors to save space: $$v\equiv [1,0,0,0,0]^T , \qquad w\equiv [0,1,0,0,0]^T,\\ v\otimes w= [0,1,0,0,...,0]^T_{25} ~.$$ It is evident that $$\lambda_1$$ acts as a straight "spin flip-flop" on the two vectors, $$v\leftrightarrow w$$, and $$\Delta(\lambda_1)~~ v\otimes w = v\otimes v + w\otimes w = \Delta(\lambda_1) ~~w\otimes v \\ =[1,0,0,0,0,0,1,0,...,0 ]^T_{25}~ .$$

Now, observe $$v\otimes w + w\otimes v$$ transforms just as above under $$\Delta(\lambda_1)$$, and is in the 15; whereas the $$v\otimes w - w\otimes v$$ in the 10 is in the kernel of $$\Delta(\lambda_1)$$; that's what makes the example too simple. In the coupled basis, it would be in the kernel of $$A_1$$.

$$A_1$$ is, of course, not trivial for su(5). Had we taken the messier $$u\equiv [0,0,1,0,0]^T$$ instead of w, we would have documented nontrivial action.

These visualization finger exercises might, or might not, be of use in your project.

• many thanks - I really appreciate it - let me digest in details. Aug 4 '21 at 16:57