The Unapologetic Mathematician

Mathematics for the interested outsider

One Complete Character Table (part 2)

Last time we wrote down the complete character table of S_3:


which is all well and good except we haven’t actually seen a representation with the last line as its character!

So where did we get the last line? We had the equation \chi^\mathrm{def}=\chi^\mathrm{triv}+\chi^\perp, which involves the characters of the defining representation V^\mathrm{def} and the trivial representation V^\mathrm{triv}. This equation should correspond to an isomorphism V^\mathrm{def}\cong V^\mathrm{triv}\oplus V^\perp.

We know that there’s a copy of the the trivial representation as a submodule of the defining representation. If we use the standard basis \{\mathbf{1},\mathbf{2},\mathbf{3}\} of V^\mathrm{def}, this submodule is the line spanned by the vector \mathbf{1}+\mathbf{2}+\mathbf{3}. We even worked out the defining representation in terms of the basis \{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2},\mathbf{3}\} to show that it’s reducible.

But what we want is a complementary subspace which is also G-invariant. And we can find such a complement if we have a G-invariant inner product on our space. And, luckily enough, permutation representations admit a very nice invariant inner product! Indeed, just take the inner product that arises by declaring the standard basis to be orthonormal; it’s easy to see that this is invariant under the action of G.

So we need to take our basis \{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2},\mathbf{3}\} and change the second and third members to make them orthogonal to the first one. Then they will span the orthogonal complement, which we will show to be G-invariant. The easiest way to do this is to use \{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2}-\mathbf{1},\mathbf{3}-\mathbf{1}\}. Then we can calculate the action of each permutation in terms of this basis. For example:


and write out all the representing matrices in terms of this basis:


These all have the required form:


where the 1 in the upper-left is the trivial representation and the 2\times 2 block in the lower right is exactly the other representation V^\perp we’ve been looking for! Indeed, we can check the values of the character:


exactly as the character table predicted.

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October 27, 2010 - Posted by | Algebra, Group theory, Representation Theory


  1. [...] Alternative Path It turns out that our efforts last time were somewhat unnecessary, although they were instructive. Actually, we already had a matrix [...]

    Pingback by An Alternative Path « The Unapologetic Mathematician | October 28, 2010 | Reply

  2. [...] a basis, we can pick . We recognize this pattern from when we calculated the invariant subspaces of the defining representation of . And indeed, is the defining [...]

    Pingback by Examples of Specht Modules « The Unapologetic Mathematician | December 28, 2010 | Reply

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