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1. The problem statement, all variables and given/known data

Decompose $\mathbb{C}^{5}$, the 5 dimensional complex Euclidean space) into invariant subspaces irreducible with respect to the group $C_{5} \cong \mathbb{Z}_{5}$ of cyclic permutations of the basis vectors $e_{1}$ through $e_{5}$.

Hint: The group is Abelian, so all the irreps are one-dimensional. Therefore, you can use the simplified form of the projection operators, with characters.

Further, try to do the same for $\mathbb{R}^{5}$, insisting that the basis vectors can only be combined with real coefficients. What is the difference between real and complex reps?

2. Relevant equations

This may be the right projection operator, unsure: $$P^{\alpha}=\frac{d_{\alpha}}{|G|} \sum_{g} \chi^{(\alpha)}(g)*O_{g}$$

3. The attempt at a solution

I am confused by the term decompose, so my attempts have been floundering. I tried to write out the character table for $\mathbb{Z}_5$ and I think I succeeded in that, but am unsure if it is needed. The hint about the projection operators served to confuse me more, although I readily understand the part about 1D irreps and Abelian. Is this asking me to construct reps (matrices) using cyclic permutations of $C_{5}$? If so, how am I supposed to use projection operators in this case to get them; This seems right however.

Any help would be wonderful.

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The "decompose" part is asking you to write $V=\mathbb{C}^5$ as $V_1\oplus\cdots\oplus V_5$ where the $V_i$ are subspaces of $V$ that are fixed by the action of the group, i.e. for all $v \in V_i$ and all $g \in \mathbb{Z}/5$ you have $gv \in V_i$. Each of those supspaces is one-dimensional of course, so you are really looking for five non-zero vectors $v_1,\ldots,v_5$ such that $gv_i$ is a scalar multiple of $v_i$. – m_t_ May 12 '12 at 19:03
The character table will help you, because it tells you how this group acts on its one-dimensional reps. Let $g$ be a generator of $\mathbb{Z}_5$. Then if $V_1=\langle v_1 \rangle$ has character $\chi$, you'll have $gv_1 = \chi(g)v_1$. You know how $g$ acts on every vector in $V$ because you know how it acts on the basis $\{e_i\}$. So to find $v_1$, write it as a linear combination of $e_i$s, and solve the equation $gv_1 = \chi(g)v_1$. Then repeat with a different character.... – m_t_ May 12 '12 at 19:16
I TeXified your post and attempted to recover the numbering by "hard-coding" the numbers. If something went wrong, I apologize. Please check that it looks all right. – Jyrki Lahtonen May 12 '12 at 19:21
For complex case you can just use the projections. In the real case you need to combine "conjugate" projections. As a hint I disclose that you should find a 2-dimensional real subspace, where the given generator acts by a 72-degree rotation, and another 2-dimensional real subspace, where the same generator acts by a 144-degree rotation. – Jyrki Lahtonen May 12 '12 at 19:23
up vote 2 down vote accepted

The generator of your group acts via the linear transformation $T:\mathbb{C}^5\rightarrow\mathbb{C}^5, (x_1,x_2,x_3,x_4,x_5)\mapsto(x_5,x_1,x_2,x_3,x_4)$. Write $\zeta=e^{2\pi i/5}$. Then the vector $$\vec{u}_j=(1,\zeta^{4j},\zeta^{3j},\zeta^{2j},\zeta^{j})$$ is an eigenvector belonging to the eigenvalue $\zeta^j$ for $j=0,1,2,3,4$. Therefore they each generate an invariant 1-dimensional (complex) subspace as described in mt_'s comments. [Edit] They can also be gotten by applying the projection operators $$ P^{-j}=\frac15\sum_{k=0}^5\chi_j^*(g^k)T^k $$ to the vector $(5,0,0,0,0)$. Here $\chi_j$ is the character $\chi_j(g^k)=\zeta^{jk}$ for all $k=0,1,2,3,4$ and $g$ is the generator of the group.[/Edit]

The real case requires a bit more work. Observe that $\zeta$ and $\zeta^4$ as well as $\zeta^2$ and $\zeta^3$ are complex conjugates of each other. Thus the complex vector spaces $U_1=\mathbb{C}\vec{u}_1\oplus\mathbb{C}\vec{u}_4$ and $U_1=\mathbb{C}\vec{u}_2\oplus\mathbb{C}\vec{u}_3$ are stable under componentwise complex conjugation, because componentwise complex conjugation simply swaps the eigenvectors. It follows that the real vector spaces $$ V_1=U_1\cap \mathbb{R}^5\qquad\text{and}\qquad V_2=U_2\cap \mathbb{R}^5 $$ are 2-dimensional (they are the eigenspaces belonging to eigenvalue 1 of the componentwise complex conjugation acting on $U_1$ and $U_2$ respectively), and also stable under the action of $T$. A calculation (see also answers to this question, in particular the link to Keith Conrad's lecture notes) shows that the matrix of $T$ with respect to the basis (over the reals) $\{\vec{u}_1+\vec{u}_4, i(\vec{u}_1-\vec{u}_4)\}$ of $V_1$ is the familiar rotation matrix. Therefore $T$ must act on $U_1$ as a rotation by 72 degrees.

Similarly the restriction of $T$ on $V_2$ is a rotation by 144 degrees.

A rotation of a real plane does not have any invariant subspaces, so we cannot refine the direct sum decomposition $$ \mathbb{R}^5=\langle(1,1,1,1,1)\rangle\oplus V_1\oplus V_2 $$ any further. Here $(1,1,1,1,1)=\vec{u}_0$.

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