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I am currently studying the book Lie Algebras in Particle Physics by Howard Georgi (2nd edition) and I couldn't understand this part:

A representation is completely reducible if it is equivalent to a representation whose matrix elements have the following form: $$\begin{bmatrix}D_1(g) & 0 & 0\ ... \\0&D_2(g) &0\ ... \\ ...\ & ...\ & ... \end{bmatrix}$$ where $D_j(g)$ is irreducible for all $j$. This is called block diagonal form.

Notation: $g$ refers to an element of a group $G$ and $D$ is a representation.

What I don't quite get is: Those diagonal elements are just numbers aren't they? Then what's with calling them irreducible representations? Going further, the author says:

A representation in block diagonal form is said to be the direct sum of the subrepresentations, $D_j$s

I am a beginner so excuse me if this is trivial. Why exactly did he say that those matrix elements are representations themselves? What am I missing?

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The elements on the diagonal $D_j(g)$, are not just numbers. They are in general matrices. That is why it is called block diagonal. As a very simple example, take SU(2): the tensor product of two SU(2) matrices, each being in the two-dimensional representation, would give you a four-dimensional matrix. Using the appropriate transformation of this matrix, one can reduce it to a block diagonal form with a three-dimensional matrix and a one-dimensional element sitting on the diagonal. The three-dimensional matrix represents the adjoint representation of SU(2) and the one-dimensional element is the singlet representation. Does this help?

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  • $\begingroup$ Thank you for answering. A follow up question: Will we still call it completely reducible if we can write it as above but the matrices $D_i$ are not irreducible? Is it even possible that they are still reducible? $\endgroup$ Oct 5 '21 at 12:44
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    $\begingroup$ @Sarthak : See some details of this flippiefanus' example in my answer here What is the symmetry of the pion triplet (π−,π0,π+)? $\color{blue}{\textbf{Example A :}}$. $\endgroup$
    – Frobenius
    Oct 5 '21 at 13:31
  • $\begingroup$ @Sarthak: Not sure I understand your question. A fully reducible case implies the block diagonal form. There are cases where it is not possible to reduce the matrix to the block diagonal form - some off-diagonal block remain non-zero. In that case, it is not fully reducible. If the matrix is fully reducible, the resulting representations obtained from the blocks on the diagonal are said to be irreducible. $\endgroup$ Oct 6 '21 at 7:54

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