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Very good article in that it taught me something I didn't know existed, and explained it very well. However, I wish the article gave an example on how to actua

Very good article in that it taught me something I didn't know existed, and explained it very well. However, I wish the article gave an example on how to actually construct a mapping between e.g. a small finite group and the actual matrices in the representation, to - sort of - get a feel for why that actually works. I suspect there must be some sort of canonical method/algorithm to get from the "multiplication table" of a finite group to each matrix in a representation, but I haven't been able to find a reference. Would anyone have pointers? Also, jumping from the group to the character table (which seem to imply that there is indeed an algorithm to compute all possible representations) without having been told how the mapping is constructed feels like a rather big mental jump (and what makes the trace of the matrices important, btw - rather than, say, the determinant?).

eigenket 6y ago
Given the multiplication table of a (finite) group there are a couple of "obvious" representations you can pick. Firstly theres always the trivial representation that just sends everything to the identity. More interestingly you can use the group elements as labels for an orthonormal basis for a vector space (e.g. your orthonormal set is {v_g for all g in G}). Then have the representation act by the usual group operation on the labels R_h[v_g] = v_{hg}. Then once you have an operation defined on a basis extending it linearly to the rest of the vector space is easy. This is (left) regular representation. https://en.wikipedia.org/wiki/Regular_representation https://en.wikipedia.org/wiki/Regular_representation Edit: Intuition for why the trace is important comes from looking at permutation matrices - the trace is the sum of the diagonal elements of the matrix and the number of 1s on the diagonal is exactly the number of points that are invariant under the permutation - e.g. the permutation [[1,0,0,0], [0,1,0,0], [0,0,0,1], [0,0,1,0]] has trace 2, and the permutation leaves the first two elements unchanged and flips the last two.
- JadeNB 6y ago
  Two remarks on this nice example: (1) The word 'orthonormal' doesn't have to be there; representations are a priori just linear actions; if they carry an invariant inner product, that is a bonus ('unitary representation'). For compact (including finite) groups, the invariant inner product is no extra information, because we could pick any (not necessarily invariant) inner product and average it with respect to Haar measure to get an invariant one. This product is unique up to a scalar for an irreducible representation, but not in general (and the left-regular representation is never irreducible for a non-trivial finite group). (2) For people who are more used to cycle notation than the matrix representation of permutations, your permutation is (3 4). In roster notation, it's 1 2 3 4 1 2 4 3
bollu 6y ago
For trace: there is a perspective that a trace is a generalized dimension counting operator. If we have a projection matrix (a matrix such that AA = A), then the trace of this matrix gives us the number of dimensions it projects onto. Indeed, requiring this property and linearity uniquely* defines trace. So, trace is the linear-algebraization of dimension counting.
sdenton4 6y ago
/Actual matrices/ : Given a particular representation R of a group, you want to break it up into a direct sum of irreducible representations, R = R0 (+) R1 (+) ... (+) Rk. Based on the character theory, you can write down this decomposition without actual matrices... But suppose you want some actual matrixes to play with. The 'direct sum' statement is the same as saying that there's a basis for R in which the group action is block diagonal, with the block elements being something from R0, R1, ..., Rk respectively. The 'concrete' requirement probably means you want to find a particular change-of-basis operation for your preferred basis of R and the block diagonal one; ie, find P such that: P R(g) P^t = R0(g) (+) ... (+) Rk(g) for every element g in the group. My lazy answer, then, is that if you have a concrete (eg, in-matrices) knowledge of R and each of the irreps, you can just solve the (many) linear equations to find P. (Remember that each g gives us another matrix equation here, which all hold simultaneously for P, so this is massively overdetermined.) Another way would be to construct projections pi from R into each of the irreducible representations, and rearrange R as pi_inverse(Image) (+) Kernel(pi) for each of these projections.

eigenket 6y ago

Given the multiplication table of a (finite) group there are a couple of "obvious" representations you can pick. Firstly theres always the trivial representation that just sends everything to the identity. More interestingly you can use the group elements as labels for an orthonormal basis for a vector space (e.g. your orthonormal set is {v_g for all g in G}). Then have the representation act by the usual group operation on the labels R_h[v_g] = v_{hg}. Then once you have an operation defined on a basis extending it linearly to the rest of the vector space is easy. This is (left) regular representation. https://en.wikipedia.org/wiki/Regular_representation https://en.wikipedia.org/wiki/Regular_representation Edit: Intuition for why the trace is important comes from looking at permutation matrices - the trace is the sum of the diagonal elements of the matrix and the number of 1s on the diagonal is exactly the number of points that are invariant under the permutation - e.g. the permutation [[1,0,0,0], [0,1,0,0], [0,0,0,1], [0,0,1,0]] has trace 2, and the permutation leaves the first two elements unchanged and flips the last two.

JadeNB 6y ago
Two remarks on this nice example: (1) The word 'orthonormal' doesn't have to be there; representations are a priori just linear actions; if they carry an invariant inner product, that is a bonus ('unitary representation'). For compact (including finite) groups, the invariant inner product is no extra information, because we could pick any (not necessarily invariant) inner product and average it with respect to Haar measure to get an invariant one. This product is unique up to a scalar for an irreducible representation, but not in general (and the left-regular representation is never irreducible for a non-trivial finite group). (2) For people who are more used to cycle notation than the matrix representation of permutations, your permutation is (3 4). In roster notation, it's 1 2 3 4 1 2 4 3

bollu 6y ago

For trace: there is a perspective that a trace is a generalized dimension counting operator. If we have a projection matrix (a matrix such that AA = A), then the trace of this matrix gives us the number of dimensions it projects onto. Indeed, requiring this property and linearity uniquely* defines trace. So, trace is the linear-algebraization of dimension counting.

sdenton4 6y ago

/Actual matrices/ : Given a particular representation R of a group, you want to break it up into a direct sum of irreducible representations, R = R0 (+) R1 (+) ... (+) Rk. Based on the character theory, you can write down this decomposition without actual matrices... But suppose you want some actual matrixes to play with. The 'direct sum' statement is the same as saying that there's a basis for R in which the group action is block diagonal, with the block elements being something from R0, R1, ..., Rk respectively. The 'concrete' requirement probably means you want to find a particular change-of-basis operation for your preferred basis of R and the block diagonal one; ie, find P such that: P R(g) P^t = R0(g) (+) ... (+) Rk(g) for every element g in the group. My lazy answer, then, is that if you have a concrete (eg, in-matrices) knowledge of R and each of the irreps, you can just solve the (many) linear equations to find P. (Remember that each g gives us another matrix equation here, which all hold simultaneously for P, so this is massively overdetermined.) Another way would be to construct projections pi from R into each of the irreducible representations, and rearrange R as pi_inverse(Image) (+) Kernel(pi) for each of these projections.