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I'm just trying to clarify my understanding of the definition of the induced representation by working with some simple examples, but I know that I must have something wrong (indeed, for the example given below, it seems like all of the representations induced from rep's of the given subgroup are reducible, but I'm certain that there should be an irreducible one). Any help would be tremendously appreciated. I suppose I can proceed via one of those examples, to have something a little more concrete (in case anyone else happens upon this question later).


Consider the so-called tetrahedral group $T:=D_2\rtimes\mathbb{Z}_3$, and its subgroup

$$D_2:=\langle c,d \ | \ c^2=d^2=e, dcd=c^{-1}\rangle=\{e_2,c,d,dc\}$$

(where the subscript on $e$ is just to distinguish the two identity elements we have). Furthermore, to set notation, we write $\mathbb{Z}=\{e_3,b,b^2\}$ and think of these as automorphisms of $D_2$, i.e.

$$b(e_2)=e_2, \quad b(c)=dc, \quad b(d)=c, \ \ \text{and} \ \ b(dc)=d.$$

Let's consider the 1D irrep $\sigma:D_2\longrightarrow GL(1,\mathbb{C})$ of $D_2$ given by $\sigma(e_2)=\sigma(dc)=1$ and $\sigma(c)=\sigma(d)=-1$. Then my understanding of how one would go about inducing a representation on all of $T$ would proceed as follows:

(1) The coset space $T/D_2$ is isomorphic to $\mathbb{Z}_3$, so we can write $D_2$, $bD_2$, and $b^2D_2$ for the cosets (i.e. using $e_3$, $b$, and $b^2$ as representatives, where we are making the identification $b\equiv (e_2,b)$, etc. for the sake of simplicity, when no confusion should arise).

(2) Let $|1\rangle$ denote a basis vector for $\mathbb{C}$, on which $\sigma$ is acting, and form a new vector space $V=\text{span}\{|e_3\rangle,|b\rangle,|b^2\rangle\}$ (i.e. use the group algebra for the coset representatives). Then one can form a new vector space $V':=V\otimes\mathbb{C}$ with basis

$$|e_3;1\rangle\equiv|e_3\rangle\otimes|1\rangle, \qquad |b;1\rangle\equiv|b\rangle\otimes|1\rangle, \quad\text{and}\quad |b^2;1\rangle\equiv|b^2\rangle\otimes|1\rangle.$$

(3) Given an arbitrary element, say for the sake of illustration $(dc,b)\in T$, determine which coset it's in. In this case

$$(dc,b)=(e_2\cdot b(c),b\cdot e_3)=(e_2,b)\cdot(c,e_3)\in bD_2.$$

(4) Then the representation $\text{Ind}^T_{D_2}(\sigma)\equiv\tilde{\sigma}:T\longrightarrow\text{GL}(3)$ can be (?) defined as, for example

$$\tilde{\sigma}(dc,b)|b^2;1\rangle=(\mathbf{R}(e_2,b)\otimes\sigma(c,e_3))|b^2;1\rangle,$$

where $\mathbf{R}$ denotes the regular representation. In other words

$$\tilde{\sigma}(dc,b)|b^2;1\rangle=|(e_2,b)\cdot(e_2,b^2);\sigma(c,e_3)1\rangle=-|(e_2\cdot b(e_2),b\cdot b^2);1\rangle=-|(e_2,e_3);1\rangle=-|e_3;1\rangle.$$

Similarly, one finds that

$$\tilde{\sigma}(dc,b)|e_3;1\rangle=-|b;1\rangle \qquad\text{and}\qquad \tilde{\sigma}(dc,b)|b;1\rangle=-|b^2;1\rangle.$$

In other words, we can write down the following matrix for $\tilde{\sigma}(dc,b)$:

$$\tilde{\sigma}(dc,b)=-\left(\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array}\right).$$

Of course, one can do similar things with all of the elements of $T$, and will end up getting either $\pm\tilde{\sigma}(dc,b)$ or the $3\times 3$ identity matrix for the other elements.

Questions: First of all, is this actually right? Is this how the induced rep. is defined/computed? Moreover, the above obtained rep. is reducible:

$$ \tilde{\sigma}(dc,b)\sim \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \exp(4\pi i/3) & 0 \\ 0 & 0 & \exp(2\pi i/3) \\ \end{array}\right), $$

and since all other matrices are this one (up to a $\pm$) or the identity, they're all diagonal in this new basis. It seems to me (and my calculations along the lines presented above) that the same is true regardless of which of the 4 irreps one chooses for $D_2$. Is there really no way to induce an irrep in this case?


EDIT: As a follow-up, in light of Derek Holt's comment below, it might be worth adding in the details of an example of an element of order 2. One such element would be $(c,e_3)$, for which I obtain:

$$ (c,e_3)=(e_2\cdot e_3(c),e_3\cdot e_3)=(e_2,e_3)\cdot(c,e_3)\in e_3D_2 $$

$$\Longrightarrow \ \tilde{\sigma}(c,e_3)|e_3;1\rangle=|(e_2,e_3)\cdot(e_2,e_3);\sigma(c)1\rangle=-|(e_2\cdot e_3(e_2),e_3\cdot e_3);1\rangle=-|e_3;1\rangle;$$

$$\quad \tilde{\sigma}(c,e_3)|b;1\rangle=|(e_2,e_3)\cdot(e_2,b);\sigma(c)1\rangle=-|(e_2\cdot e_3(e_2),e_3\cdot b);1\rangle=-|b;1\rangle;$$

$$\qquad \ \ \tilde{\sigma}(c,e_3)|b^2;1\rangle=|(e_2,e_3)\cdot(e_2,b^2);\sigma(c)1\rangle=-|(e_2\cdot e_3(e_2),e_3\cdot b^2);1\rangle=-|b^2;1\rangle;$$

thus

$$ \tilde{\sigma}(c,e_3)= -\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right). $$

Similarly, all other elements of order 2 will yield $\pm I_3$ ($+$ for $(e_2,e_3)$ and $(dc,e_3)$; $-$ for $(c,e_3)$ and $(d,e_3)$). All elements of order 3 will give $\pm\tilde{\sigma}(dc,b)$. So, the representation will definitely be reducible, and precisely the same story will play out had we chosen any of the other 3 available choices of $\sigma$ -- all 4 cases give a reducible representation. Since Derek Holt seems to be saying that the 3D irrep of $T$ can be obtained by induction, something must be wrong with the way that I'm computing these matrices/the definition of the induced representation I'm using.

Where am I going wrong?

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  • $\begingroup$ Induced reps are usually reducible; that shouldn't be surprising. $\endgroup$ – Qiaochu Yuan Nov 22 '17 at 8:44
  • $\begingroup$ Yeah, it's not so much that it is reducible in this case, it's that it seems like it always is (for $T$ and $D_2$) -- this method doesn't seem to give you any irreps for $T$. I suppose that just feels odd, since $D_2$ only has 1D irreps and there are 3 coset representatives, so the induced rep will always be 3D... and $T$ does, in fact, have an irreducible 3D rep (since it has 4 conjugacy classes and 12 can only be written as the sum of 4 perfect squares in one way: $1^2+1^2+1^2+3^2$). Is this 3D irrep not obtainable via the induction method? Or did I just mess something up in my computations? $\endgroup$ – OperaticDreamland Nov 22 '17 at 9:08
  • $\begingroup$ This induced representation should be irreducible. The elements of order $3$ should map to permutation matrices but with entries $\pm 1$ (as you have computed), and those of order $2$ should map to diagonal matrices with $\pm1$ on the diagonal. The result is a so-called monomial representation. There is no basis in which all of these matrices are simultaneously diagonal. $\endgroup$ – Derek Holt Nov 22 '17 at 9:24
  • $\begingroup$ In that case, something must be wrong with my understanding of how the induced representation is defined. I'm only getting three matrices showing up: the one above, it's square, and the identity matrix. These are definitely simultaneously diagonalizable. Any idea what is wrong in the way I'm computing them? $\endgroup$ – OperaticDreamland Nov 22 '17 at 13:29
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    $\begingroup$ Sorry but I have not time to check your calculation in detail. The diagonal entries on the matrix defined by applying your induced representation to $(c,e_3)$ should be $-1$, $-1$ and $1$ rather than $-1$ three times. They should be arising from applying $\sigma$ to the $c$, $cd$ and $d$ which are the three conjugates of $c$ under the conjugation action of $1$, $b$ and $b^2$ on $c$. The fact that $\sigma(c)$ is occurring in your calculation of each of these three entries looks suspicious to me, and may be the source of the error. $\endgroup$ – Derek Holt Nov 24 '17 at 18:29
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Just in case anyone happens across this, I figured out my error -- I was doing things in the wrong order. Rather than first determining which coset one is dealing with, and then doing the relevant multiplication, one should be doing the multiplication and determining which coset the result lies in instead.

For example, in the case of $\tilde{\sigma}(c,e_3)$ given in the edit above, one should have:

$$ \quad \ \ \ \tilde{\sigma}(c,e_3)|e_3;1\rangle=|(c,e_3)\cdot(e_2,e_3);1\rangle=|(e_2,e_3)\cdot(c,e_3);1\rangle=|(e_2,e_3);\sigma(c)1\rangle=-|e_3;1\rangle; $$

$$ \tilde{\sigma}(c,e_3)|b;1\rangle=|(c,e_3)\cdot(e_2,b);1\rangle=|(e_2,b)\cdot(d,e_3);1\rangle=|(e_2,b);\sigma(d)1\rangle=-|b;1\rangle; $$

$$ \qquad\tilde{\sigma}(c,e_3)|b^2;1\rangle=|(c,e_3)\cdot(e_2,b^2);1\rangle=|(e_2,b^2)\cdot(dc,e_3);1\rangle=|(e_2,b^2);\sigma(dc)1\rangle=|b^2;1\rangle; $$

hence, one obtains the matrix for $(c,e_3)$ as

$$ \tilde{\sigma}(c,e_3)=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right), $$

as one should. Similarly for the other computations. This is, indeed, the 3D irrep of $T$ mentioned by Derek Holt in his comments above.

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