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  Physical interpretation of the rank of $SU(N)$ ?

+ 2 like - 0 dislike
I think I have understood that the group $SU(N)$ can be represented by $N\times N$ matrices with unit determinant, and why it has $N^2 -1$ generators. However I dont see why the rank of $SU(N)$ is $N-1$, how can this generally be proved? Also, why is it that the rank of $SU(N)$ gives the number of operators in the algebra that can simultaneously be diagonalized and what does it mean from a physics point of view?
asked Dec 30, 2014 in Mathematics by Dilaton (6,240 points) [ revision history ]

2 Answers

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The rank is the dimension of a maximal abelian subgroup (called the torus). In a matrix group, one torus is given by the diagonal matrices. For $SU(N)$, Since the determinant has to be one, there are $N-1$ free $U(1)$ parameters. It's easy to check that this is maximal.

It's also an easy theorem in linear algebra that two diagonalizable matrices can be simultaneously diagonalized iff they commute.

One can devise many physical situation where the rank matters. Suppose some adjoint matter gets a vev. Then the gauge field splits into a number of photons. The number is the rank.
answered Dec 30, 2014 by Ryan Thorngren (1,925 points) [ revision history ]
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I understand the rank $N$ of a gauge group as the degrees of freedom of the fundamental representation of the gauge group. The easiest example is to consider the color group $SU(3)$ of the standard model and then the fundamental representation which is given by the quark wave function. Then, this is/transforms in the ${\bf{3}}$, and equivalently the anti quarks in the ${\bar{\bf{3}}}$, right? This indeed gives us the degrees of freedom. Similar argument holds for the adjoint representation.
answered Jan 4, 2015 by Outlander (95 points) [ revision history ]
edited Jan 5, 2015 by Outlander

This only works for $SU(N)$ but not for $SO(n)$ with $n>2$.

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