# Are the irreps of SO(n) orthogonal?

This seems like a trivial question, but I'm wondering if irreducible representations automatically inherit the properties of the group that they represent. Specifically, if I take an irrep of SO(4) over the complex numbers will it be orthogonal; i.e. if R is an irrep of SO(4) with complex matrix elements does $R^{T}R=I$ hold?

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@Thomas: Well, I'm not exactly sure what $X^T$ would mean. But if $X \in SO\left(4\right)$, and $R\left(X\right)$ is an irrep of $X$, then certainly $R^{-1}\left(X\right) = R\left(X^{-1}\right)$. I suppose the real question is whether or not $R\left(X^{-1}\right) = R^T\left(X\right)$. – okj Mar 6 '13 at 19:31
Well, the obvious counterexample is, when $n=2$ and we use the 1-dimensional complex representation of $SO(2)$ $$R:\pmatrix{\cos\phi&\sin\phi\cr-\sin\phi&\cos\phi\cr}\mapsto e^{i\phi}.$$ All the 1x1 matrices are there own transposes, but not there own inverses. – Jyrki Lahtonen Mar 6 '13 at 19:52
And in general the question is a bit odd, because the matrices $R(X)$ of a representation $R:SO(n)\to GL(V)$, $V$ a complex vector space, depend on the choice of basis for $V$. Orthogonality on the other hand is not preserved under change of basis, so the answer depends on the choice of basis. It is well known that we can always arrange the matrices representing a compact Lie group to be unitary by picking the basis intelligently. – Jyrki Lahtonen Mar 6 '13 at 19:57
@JyrkiLahtonen: So I guess it all comes down to what the OP means by $R^T$? If it is defined by $R^T(X) = R(X^T)$, then it is true, right? – Thomas Mar 6 '13 at 19:59
I don't think you can define another representation $R^T$ by setting $R^T(X)=R(X^T)$. As taking the transpose is an anti-automorphism of $SO(n)$, that is not a representation! But, yeah, you're right, the OP should clarify what s/he means by $R^T$. I thought that the question is whether $R(X)^T=R(X)^{-1}$ for all $X\in SO(n)$? – Jyrki Lahtonen Mar 6 '13 at 20:02

[Fleshing out my comments a bit, as more details were requested.]

If $\rho:G\to GL(V)$ is a complex representation of a matrix group $G$, then many questions like this are made a bit unnatural by the fact that there may be no canonical basis for $V$. To turn the linear mappings $\rho(g), g\in G$ into matrices $R(g)$ we need to specify a basis for $V$. Even the defining representation suffers from this. Consider the group $SO(2)$ of rotations of the real plane. If we use the natural basis $e_1=(1,0)^T, e_2=(0,1)^T$, rotation $Rot(\phi)$ by the angle $\phi$ is represented by the well known orthogonal matrices $$Rot(\phi)\mapsto\left(\begin{array}{rr}\cos\phi&-\sin\phi\\ \sin\phi&\cos\phi\end{array}\right).$$ However, if we treat this as a complex representation, and use the basis $f_1=(1,i)^T, f_2=(1,-i)^T$ instead, then the matrix representing $Rot(\phi)$ becomes $$Rot(\phi)\mapsto\left(\begin{array}{ll}e^{-i\phi}&0\\ 0&e^{i\phi}\end{array}\right),$$ as $\{f_1,f_2\}$ is an eigenbasis for all rotations. We see that as a complex representation (after extension of scalars) the defining two-dimensional real representation lost its irreducibility: the vectors $f_1$ and $f_2$ both span a subrepresentation. Note that neither of these component representations have the property that all the matrices of $\rho(g)$ would have determinants equal to $1$.

When we construct another representation of $G$, the lack of a natural basis is, of course, an even more pressing concern.

This cloud has a silver lining though. For any complex representation of a compact Lie group (including all finite groups) $\rho: G\to GL(V)$ we can equip $V$ with an inner product with the property that $\langle \rho(g)x,\rho(g)y\rangle=\langle x,y\rangle$ for all $x,y\in V$ and all $g\in G$. We achieve this by selecting any inner product $(,)$ on $V$, and then calculating the average over the action of $G$: $$\langle x,y\rangle:=\int_G(\rho(g)x,\rho(g)y)\,d\mu,$$ where $d\mu$ is the Haar measure of $G$. Then if we pick any basis of $V$ that is orthonormal w.r.t. this special inner product, the matrices of $\rho(g)$ will all be unitary. Similarly, if $V$ is real vector space, we can make all the matrices $\rho(g)$ orthogonal.

So a lot revolves on selecting a nice basis for $V$. A central theme in the representation theory of simple algebraic groups is our ability to select such a basis for any irreducible f.d. representation such that all diagonal matrices are represented by diagonal matrices with respect to this basis. I'm not sure how much of that bit survives in the simple Lie group side, because you would have to first alter the defining representation in such a way that the elements of a torus are all diagonal?? But this shows that questions of this kind may be very interesting.

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Thanks a lot! That's very helpful. – okj Mar 7 '13 at 18:17