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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:

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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 $SU(5)$.

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  • $\begingroup$ Crossposted to math.stackexchange.com/q/4214397/11127 $\endgroup$
    – Qmechanic
    Commented Aug 2, 2021 at 14:25
  • $\begingroup$ Yes thanks- I thought math and physics people could provide different types of thinkings - which indeed people do. :) $\endgroup$ Commented Aug 2, 2021 at 14:26

2 Answers 2

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  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).

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  • $\begingroup$ thanks so much for this +1! $\endgroup$ Commented Aug 1, 2021 at 19:03
  • $\begingroup$ But sorry in which equations do you obtain the 10-dimensional matrix representations of π‘†π‘ˆ(5) and the 15-dimensional matrix representations of π‘†π‘ˆ(5) ? $\endgroup$ Commented Aug 1, 2021 at 19:04
  • $\begingroup$ This part is nice "𝑒(5)=𝑒(5)π΄βŠ•π‘’(5)𝑆(9) of anti-Hermitian 5×5 matrices, which separates into a 10-dimensional subspace 𝑒(5)𝐴 of real antisymmetric matrices, and a 15-dimensional subspace 𝑒(5)𝑆 of imaginary symmetric matrices." but we hope to make these rank-10 and rank-15 matrices explicitly $\endgroup$ Commented Aug 1, 2021 at 22:30
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    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Aug 2, 2021 at 14:55
  • $\begingroup$ Thanks so much for the great answer --- could you clarify: " adjoint representation acts on the Lie algebra 𝑒(5), but it does not respect the splitting $𝑒(5)=𝑒(5)_π΄βŠ•π‘’(5)_𝑆$." What do we learn from the properties here? (I admire you including the discussions, points by points, so I could easily follow.) $\endgroup$ Commented Aug 26, 2021 at 15:43
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Unfortunately, despite my near-promise in the question you quote, I don't know of a source that computes these bulky sets of 24 10Γ—10 and 15Γ—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 5Γ—15 generator in your non-standard normalization, $A_1$ for the corresponding 10Γ—10 one, and $S_1$ for the 15Γ—15 one. But, alas!, I won't even get to computing those, but just the reducible 25Γ—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 5Γ—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 25Γ—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.

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  • $\begingroup$ many thanks - I really appreciate it - let me digest in details. $\endgroup$ Commented Aug 4, 2021 at 16:57

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