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1) The tensor product of two matrices is define by

[tex] A \otimes B =\left( {\begin{array}{cc}

a_{11}B & a_{12}B \\

a_{21}B & a_{22}B \\

\end{array} } \right)

[/tex]

for the 2x2 case with obvious generalisation to higher dimensions.

The tensor product of two groups is defined by

[tex] G \otimes H = \ { (g,h) \vert g \epsilon G , h\epsilon H \ }

[/tex]

My question is this: Does the tensor product, defined in the first equation, of two matrix representations of groups A and B form a representation of the tensor product of the two groups as defined in the second equation? And if so, why?

A general element of the group AxB is (a,b), with the group operation defined in the obvious way ie

[tex] (a_1 , b_1) * (a_2 , b_2) = (a_1 a_2 , b_1 b_2) [/tex]

My first instinct to form a matrix representation of the group AxB would be to take matrix reps of A and B and make a block diagonal matrix out of them. Then if I multiplied together two such matrices I would get exactly the multiplication law described above.

However, the tensor product of matrix reps of A and B gives something completely different. It is not clear to me that such a matrix should be a rep of the group AxB, or that it should obey the above multiplication law if we use matrix multiplication.

2) Assuming that the tensor product of matric reps of groups A and B gives a matrix rep of the group AxB, then:

If the matrix reps of A and B are in the the fundamental representation, does the matrix I get by taking their tensor product live in the fundamental representation of AxB?

Finally...

3) This all came from a statement in a book Im reading at the moment, namely:

"The electromagnetic current belongs to the singlet and adjoint representation of SU(N)"

where the eleectromagnetic current is

[tex] j_{\mu} = \psi_a \gamma_{\mu} \psi_b [/tex]

where a and b are flavour indices transforming under the fundamental represenation of SU(N). It is not clear to me why this object should be in the "singlet and adjoint representation" of SU(N).

Thanks in advance.