# How to determine a basis for invariant subspaces transforming in the irreps of a finite group?

I have what I think is a relatively simple problem, at least for the case at hand it should be (my group is only has 12 elements). I provide some specific details below for my problem, but skip to below if you don't care about geometric point groups, the question does not rely explicitly on this particular example.

Background

The group $$D_{3d}$$ is the symmetry group of a triangular anti-prism, which looks like this:

I have highlighted the $$C_3$$ axis (120 degree rotations) in black and the three $$C_3$$ axes (120 degree rotation) in blue. The red axes aren't important for this discussion. The set of all of the symmetries are the identity $$\mathfrak{e}$$, inversion through the midpoint, $$\mathfrak{i}$$, two $$C_3$$ rotations about the black axis, three $$C_2$$ rotations (one for each blue axis), two roto-inversions $$S_6$$ ($$C_3$$ rotation followed by inversion), and three mirror planes $$\sigma_d$$ ($$C_2$$ rotation followed by inversion). There are twelve total elements of the group.

I can form a 6-dimensional representation of this group which is the set of permutation matrices for the six outer corners, which looks like

I can then write them all as permutations, $$\begin{array}{|c|c|c|} \hline \text{name} & \text{order} & \text{permutation} \\\hline \mathfrak{e} & 1 & \text{no permutation} \\\hline \mathfrak{i} & 1 & (14)(25)(36) \\\hline C_3 & 2 & { (135)(246)\\ (531)(642) } \\\hline C_2 & 3 & { (14)(23)(56)\\ (25)(16)(34)\\ (36)(12)(45) } \\\hline \sigma_d & 3 & { (26)(35)\\ (13)(46)\\ (15)(24) } \\\hline S_6 & 2 & { (123456)\\ (654321) } \\\hline \end{array}$$

This representation is reducible, and we can calculate how it decomposes using the character table of $$D_{3d}$$

$$\begin{array}{|c|c|c|c|c|c|c|} \hline & \mathfrak{e} & 2C_3 & 3C_2 & \mathfrak{i} & 2S_6 & 3\sigma_d \\ \hline A_{1g} & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline A_{1u} & 1 & 1 & 1 & -1 & -1 & -1 \\ \hline A_{2g} & 1 & 1 & -1 & 1 & 1 & -1 \\ \hline A_{2u} & 1 & 1 & -1 & -1 & -1 & 1 \\ \hline E_g & 2 & -1 & 0 & 2 & -1 & 0 \\ \hline E_u & 2 & -1 & 0 & -2 & 1 & 0 \\ \hline \end{array}$$ so that this representation $$(R)$$ decomposes as a sum of irreps $$r$$ (whose labels appear in the left column), $$R = \oplus_{r}\, n_r \,r$$ The numbers of time each irrep appears are found from the formula $$n_r = \frac{1}{12} \sum_{g\in D_{3d}} \chi_r^*(g) \chi_R(g)$$ where $$g$$ are the group elements, $$\chi_r(g)$$ is its character from the character table, and $$\chi_R(g)$$ is the trace (character) of the 6-dimensional representation matrices. I find from this that the 6-dimensional permutation representation decomposes as $$R = A_{1g}\oplus A_{2u} \oplus E_g \oplus E_u$$

Question

Now my question is: how do I obtain the linear combinations of corners which transform in the irreps of dimension greater than 1?

My attempt

More specifically, I start with the free vector space whose basis are the corners labelled $$i$$, so a vector takes the form $$\vert \psi \rangle = \sum_{i=1}^6 c_i \vert i \rangle \quad \, c_i \in \mathbb{R}$$ I want to find a basis for each of the invariant subspaces. This is easy to do for the 1d irreps, because I know that they act by simply multiplying by the respective characters: construct a generic linear combination of the basis vectors as above and ensure that under each permutation it changes by the appropriate sign.

But I cannot figure out a general way to do this for the 2d irreps, since I don't explicitly have the $$2\times 2$$ matrices for those irreps. My attempt to do this was by explicitly constructing the transformation: for a given permutation $$p\in R$$ which send $$i\to p(i)$$, I need to enforce that $$\begin{pmatrix} \sum_i a_{p(i)} \vert i \rangle \\[10pt] \sum_i b_{p(i)} \vert i \rangle \end{pmatrix} = \begin{pmatrix} A & B \\[10pt] C & \chi_{r}(p)-A \end{pmatrix} \begin{pmatrix} \sum_i a_{i} \vert i \rangle \\[10pt] \sum_i b_{i} \vert i \rangle \end{pmatrix} \quad \forall \, p \in R$$ where $$r$$ is either $$E_g$$ or $$E_u$$, and the vectors $$\sum_i a_i \vert i \rangle$$ and $$\sum_i b_i \vert i \rangle$$ ideally form an orthonormal basis for the invariant subspace (of course the $$a_i$$ and $$b_i$$ are only determined up to a rotation within the subspace). But I don't know what the 2d representation matrix, i.e. what the constants $$A,B,C$$ are for the different symmetry elements. It seems this is not a good way to go about finding the basis vectors, is there a straightforward way to construct basis vectors for different subspaces?

Question

Now my question is: how do I obtain the linear combinations of corners which transform in the irreps of dimension greater than 1?

As I understand it, chemists/physicists say a vector $$v$$ (from a vector space $$V$$ on which a group $$G$$ acts by a linear representation) "transforms as" an irreducible representation $$W$$ if the subrepresentation of $$V$$ generated by $$v$$ (that is, the subspace of all linear combinations of the vectors $$\rho_V(g)v$$ for $$g\in G$$) is equivalent, as a representation, to the irrep $$W$$.

Mathematicians might say $$v$$ "cyclically generates" a subrep equivalent to the irrep $$W$$.

Fun fact: every nonzero vector in an irrep is a cyclic generator. So if you have a representation $$V$$ with an unknown subirrep $$W$$, to find a cyclic generator of $$W$$ it suffices to find any nonzero element of it.

Brief tangent: Maschke's theorem says any complex representation of a finite group decomposes as a direct sum of subirreps. In general, these subirreps are not unique. However, if in such a decomposition one were to group together all subirreps equivalent to a given irrep $$W$$, their direct sum is uniquely determined. It is called the "isotypical component" of $$V$$ (of type $$W$$). Every subirrep of $$V$$ equivalent to $$W$$ is contained in this isotypical component, and conversely the component is the sum of these subirreps.

Because Artin-Wedderburn says the group algebra $$\mathbb{C}[G]$$ is isomorphic to $$\bigoplus\mathrm{End}(U)$$ (or a direct sum $$\bigoplus_i M_{d_i}(\mathbb{C})$$ of matrix algebras if you prefer coordinates), we ought to be able to find an isotypical projector $$e_W\in\mathbb{C}[G]$$ which annihilates vectors in irreps $$U$$ inequivalent to $$W$$ while it fixes vectors in the irrep $$W$$. (This correspond to the element of $$\bigoplus\mathrm{End}(U)$$ which is $$0_U$$ on $$U\ne W$$ and $$1_W$$ on $$W$$.)

The isotypical projector associated to an irrep $$W$$ with character $$\chi_W$$ is

$$e_W=\frac{\dim W}{|G|}\sum_{g\in G}\overline{\chi_W(g)} g.$$

So, let's take an element of your 6D rep, say $$|1\rangle$$, and project it down to the $$E_g$$ subrep for instance. I've extended your table to include $$g|1\rangle$$ and $$\chi_V(g)$$ for all your permutations $$g$$:

$$\begin{array}{|c|c|c|c|c|} \hline \text{name} & \text{order} & g & g|1\rangle & \chi_{E_{\large g}}(g) \\\hline \mathfrak{e} & 1 & () & |1\rangle & 2 \\\hline \mathfrak{i} & 1 & (14)(25)(36) & |4\rangle & 2 \\ \hline C_3 & 2 & { (135)(246)\\ (531)(642) } & {|3\rangle \\ |5\rangle} & -1 \\ \hline C_2 & 3 & { (14)(23)(56)\\ (25)(16)(34)\\ (36)(12)(45) } & {|4\rangle \\ |6\rangle \\ |2\rangle} & 0 \\\hline \sigma_d & 3 & { (26)(35)\\ (13)(46)\\ (15)(24) } & { |1\rangle \\ |3\rangle \\ |5\rangle} & 0 \\\hline S_6 & 2 & { (123456)\\ (654321) } & {|2\rangle \\ |6\rangle} & -1 \\\hline \end{array}$$

Then the projection of $$|1\rangle$$ onto the $$E_g$$ subrep is

$$2|1\rangle-|2\rangle-|3\rangle+2|4\rangle-|5\rangle-|6\rangle$$

or as a coordinate vector, $$(2,-1,-1,2,-1,-1)$$. You can similarly project $$|2\rangle,|3\rangle,|4\rangle,|5\rangle,|6\rangle$$ down to find a spanning set for the $$E_g$$ subirrep, and toss out extraneous vectors to get a basis.

If you've done lots of calculations with small reps before, you might recognize $$2,-1,-1$$...

The above is the most general way to go (assuming you have the character table of your group at hand). Sometimes you can get away with using special knowledge of your group and the representation $$V$$ to decompose it though. In this case, $$V$$ is a permutation representation, so we can consider $$G$$'s permutation action on $$\{1,\cdots,6\}$$. Notice, clearly, $$\{\{1,3,5\},\{2,4,6\}\}$$ is a stable partition. Taking inspiration from the standard 2D rep of the symmetric group of degree three, we can construct a subrepresentation of $$V$$ consisting of all linear combinations $$\sum x_i|i\rangle$$ for which $$x_1+x_3+x_5=x_2+x_4+x_6=0$$ and $$x_i=x_{i+3\bmod 6}$$ (note $$i$$ and $$i+3\bmod6$$ are antipodal vertices of the antiprism). One has to "copy and paste" the values $$x_1,x_3,x_5$$ to the other terms $$x_2,x_4,x_6$$ because our permutations can switch the two sides of the antiprism.

Or, one could instead use the condition $$x_{i+3\bmod 6}=-x_i$$, which presumably defines the $$E_u$$ subrep. One could also look at the stable partition $$\{\{1,4\},\{2,5\},\{3,6\}\}$$ instead, which using the idea of copy-pasting coordinates with $$\pm$$ I suspect gives the $$A_{1g}$$ and $$A_{2u}$$ subirreps.

• Thanks! I used that same projector formula (strictly speaking, shouldn't it be $\rho(g)$ and not $g$?) and was able to find a basis, including that vector you showed there, but I did not have the extra insight from your last two paragraphs. I'm still learning this stuff so I will have to read carefully and think about it for a while, I am not familiar with either Maschke's theorem or Artin-Wedderburn, but this is really helpful. Could you clarify what you mean by "stable partition"?
– Kai
Commented Sep 25, 2020 at 14:17
• (a) Yes strictly speaking it should be $\rho(g)v$ not $gv$, but in representation theory we often abbreviate the former as the latter. (b) I mean if you apply any of the permutations $g$ (coming from your symmetry group) to the partition, it's still the same partition. Also it occurs to me I forgot to multiply by the normalizing constant $\dim W/|G|$ in my projections, but the result will be the same up to scaling so it doesn't matter.
– anon
Commented Sep 25, 2020 at 19:29