Complex and Real Representations, their differences by decomposition

[aquila457]

Homework Statement

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?

Homework Equations

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

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.

 

[mighty2000]

Decomposing a vector space means breaking it down into smaller, simpler subspaces. In this case, we are looking for invariant subspaces, which are subspaces that are preserved by the group action. In this problem, the group C_{5} is acting on the vector space \mathbb{C}^{5} by permuting the basis vectors. Invariant subspaces are subspaces that remain unchanged under this action, so they are spanned by vectors that are fixed by the group action.

To find these invariant subspaces, we can use the projection operators, as suggested in the hint. These operators can be written as P^{\alpha}=\frac{1}{|G|} \sum_{g} \chi^{(\alpha)}(g)*O_{g}, where \alpha is the irrep label, |G| is the order of the group, \chi^{(\alpha)}(g) is the character of the irrep \alpha at the element g, and O_{g} is the operator that corresponds to the group element g. In this case, since the group is Abelian, the irreps are one-dimensional and the characters are simply +1 or -1, depending on whether the group element fixes the vector or not. We can use these projection operators to find the invariant subspaces of \mathbb{C}^{5} by projecting onto the eigenvectors of the group elements.

For example, for the first basis vector e_{1}, the projection operator would be P^{(1)}=\frac{1}{5} \left( 1+1+1+1+1 \right)=\frac{5}{5}=1. This means that the subspace spanned by e_{1} is invariant under the group action, and thus is an invariant subspace. Similarly, for the second basis vector e_{2}, the projection operator would be P^{(1)}=\frac{1}{5} \left( 1-1+1-1+1 \right)=0, which means that the subspace spanned by e_{2} is not invariant and can be decomposed into smaller invariant subspaces.

To find the difference between real and complex irreps, we can repeat this process for the vector space \mathbb{R}^{5}. In this case, since the basis vectors can only be combined with real coefficients, the projection