Induced Representations

peccavi_2006 · **Posted:** Thu, 17 May 2012 01:03:06 UTC

It's an easy example (better to start off with something simple) but I'm trying to find
$V := \mathrm{Ind}_{S_{2}}^{S_{3}} 1$ , where 1 is the trivial representation on $S_{2}$ .

So, I want to use Froebenius Reciprocity. Then
$< \chi_{V}, \chi_{1} > _{S_{3}} = < \chi_{1}, \chi_{1}|_{S_{2}} > _{S_{2}} = 1$ ; doing a similar calculation with the character of the sign representation gives 0, so therefore
$V = 1 \oplus nU$ , where U is an irreducible 2-dimensional representation on $S_{3}$ and n is the multiplicity.
Now, the dimension of V is equal to $[S_{3}: S_{2}] dim(1)$ ; but the dimension of the trivial representation is 1, $S_{2} \cong C_{2}$ and $[S_{3}: S_{2}] = |S_{3}| / |C_{2}| = 6/2 = 3$ , so dimV = 3 and hence n=1, i.e. $V = 1 \oplus U$ .
Of course, to show this explicitly I would have to calculate the restriction of U on $S_{2}$ , show that it is equal to $1 \oplus sign$ and then go back to characters, but without doing that bit (im still working through the restriction of U) does it look like a semi-convincing argument?

outermeasure · **Posted:** Thu, 17 May 2012 07:59:56 UTC

peccavi_2006 wrote:

It's an easy example (better to start off with something simple) but I'm trying to find
$V := \mathrm{Ind}_{S_{2}}^{S_{3}} 1$ , where 1 is the trivial representation on $S_{2}$ .

So, I want to use Froebenius Reciprocity. Then
$< \chi_{V}, \chi_{1} > _{S_{3}} = < \chi_{1}, \chi_{1}|_{S_{2}} > _{S_{2}} = 1$ ; doing a similar calculation with the character of the sign representation gives 0, so therefore
$V = 1 \oplus nU$ , where U is an irreducible 2-dimensional representation on $S_{3}$ and n is the multiplicity.
Now, the dimension of V is equal to $[S_{3}: S_{2}] dim(1)$ ; but the dimension of the trivial representation is 1, $S_{2} \cong C_{2}$ and $[S_{3}: S_{2}] = |S_{3}| / |C_{2}| = 6/2 = 3$ , so dimV = 3 and hence n=1, i.e. $V = 1 \oplus U$ .
Of course, to show this explicitly I would have to calculate the restriction of U on $S_{2}$ , show that it is equal to $1 \oplus sign$ and then go back to characters, but without doing that bit (im still working through the restriction of U) does it look like a semi-convincing argument?

Well, you can do it that way. In this case, it is much faster to just work out $\chi_V$ directly from the formula $\displaystyle\chi_V(g)=\sum_{x\in T}1_{S_2}(g^x)$ (where

is a right transversal of S_2<S_3

) to show $\chi_V=\mathop{\mathrm{fix}}$ , since there is only three conjugacy classes of S_3

.

But for more complicated examples, yes, Frobenius reciprocity is a great tool to work out the decomposition.

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Induced Representations

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