Antisymmetric powers of $SO(n)$ representation. I am particularly interested for $SO(3)$.
Let us say that I start with the natural/defining $3$-real-dimensional vectorial representation of $SO(3)$ and I choose the generator of rotation in the $1$-$2$ plane as my Cartan subalgebra. So I have $3$ weight vectors with weights $1$, $0$, $-1$ w.r.t. this generator.
Then is there a way to see what will be the weights under this chosen Cartan of $SO(3)$ of the weight vectors in the $m$-fold antisymmetrization of this $3$-dimensional vectorial representation?


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*It would be great if possible someone can explain the corresponding general result for $SO(n)$, $U(n)$ and $SU(n)$ where too I think the same question can be asked. 

 A: In general the multiset of weights of the $m$-th exterior power of a representation $V$ (I take it that this is what you cal $m$-fold anti-symmetrization) is obtained by taking the sums of all $m$-subsets of weights of $V$. So in the $SO(3)$ case the second exterior power of the defining representation has weights $1+0=1$, $1+-1=0$, and $0+-1=-1$ (it is isomorphic to the defining representation itself), the third exterior power has weight $1+0+-1=0$ (it is the trivial representation), and all further exterior powers are the zero representation (no weights).
You can similarly work out the other cases. For $U(n)$ and $SU(n)$, the exterior powers of the defining representation traverse all $n$ "fundamental representations" and then become zero. For $SO(n)$, one basically traverses the fundamental representations once and then again in reverse order (because the defining representation is self-dual), but there are some exceptions near the turning point, where the (fundamental) spin representations are not obtained, but others are in their place. However, the above description of the set of weights is valid in all cases.
A: In the particular case of $SO(3)$, one can understand what is happening by general principles.
If $V$ is a $3$-dimensional vector space, then $\wedge^2 V$ is again $3$-dimensional, while $\wedge^3 V$ is one-dimensional, and there is a canonical pairing $V \times \wedge^2 V \to \wedge^3 V$, which induces a natural identification
$\wedge^2 V \cong V^{\vee} \otimes \wedge^3 V$.
Now take $V$ to be the standard $3$-dimensional rep'n of $SO(3)$.  Since an endomorphism of $V$ acts on $\wedge^3 V$ through its determinant, and the elements of $SO(3)$ have determinant $1$ by definition, we see that $\wedge^3 V$ is the trivial rep'n of $SO(3)$.  So we obtain an isomorphism of representations $\wedge^2 V \cong V^{\vee}$.  
But the standard inner product identifies $V$ with $V^{\vee}$, and this identification is compatible with the $SO(3)$-actions (by definition of an orthogonal matrix).
Conclusion:
We have $\wedge^2 V \cong V$ as $SO(3)$-rep's, while $\wedge^3 V$ is the trivial rep'n of $SO(3)$.
