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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?

  • 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.
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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.

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  • $\begingroup$ Thanks for the reply. Can you say something more about what definition of "fundamental" representation of $U(n)$ or $SU(n)$ that you are using? For such higher rank groups one would have to keep track of weights under multiple Cartans. So can you say something along the lines of what happens to the weights under the various Cartans as one takes higher antisymmetric powers? $\endgroup$
    – user6818
    Jul 29, 2012 at 16:29
  • $\begingroup$ What I call a fundamental representation (I'm not sure the term is correct) is an irreducible representation with highest weight a fundamental weight: one whose evaluation on one of the simple coroots is $1$, and on all other simple coroots it is $0$. The number of fundamental weights is equal to the rank of the group, but they might be weights only for the universal cover of a group, and in particular for $SO(n)$ one or two fundamental weights correspond to spin representations not of $SO(n)$ but of the spin group $\mathrm{Spin}(n)$ ccovering it. $\endgroup$
    – Marc van Leeuwen
    Jul 29, 2012 at 18:19
  • $\begingroup$ As for higher rank groups, one still needs weights only for a single Cartan (since all Cartan subgroups of a compact group are conjugate), but such a weight has as many components as the rank of the group (and the CSA has as many independent generators). $\endgroup$
    – Marc van Leeuwen
    Jul 29, 2012 at 18:27
  • $\begingroup$ I guess we are having a difference in terminology. I guess by a "single Cartan" you mean a set of rank number of generators for a choice of Cartan sunalgebra. I guess in a fundamental representation in some convention it should be possible to say that the Dynkin index of the highest weight is $(1,0,0,0,..)$..right? And what is your definition for "anti-fundamental"? Is there a reference which gives a hand-on demonstration of taking higher exterior powers of fundamental and antifundamental representation of $SO(n)$, $U(n)$ and $SU(n)$ and how the weights under the Cartans change? $\endgroup$
    – user6818
    Jul 30, 2012 at 19:38
  • $\begingroup$ By a "single Cartan" I mean a single Cartan subgroup (or maybe Cartan subalgebra if the context requires that). The notion of weight for a Cartan subgroup or algebra is defined without using a choice of basis or set of generators. However, once a choice of positive roots is made, and therefore of simple (co)roots, the simple coroots provide a natural set of "coordinates" on the weights, and a fundamental weight is one that has a single such coordinate $1$ and all other $0$. In terms of a $W$-invariant inner product, the coroot for $\alpha$ is $\lambda\mapsto(\lambda,\alpha)/(\alpha,\alpha)$. $\endgroup$
    – Marc van Leeuwen
    Jul 31, 2012 at 7:47
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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 $SO(3)$.

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  • $\begingroup$ I guess by $V^{v}$ you mean the dual of $V$..right? So the weights under the antisymmetrization of $SO(3)$ seems to behave distinctly differently than that of $U(n)$ or $SU(n)$ where the weights just keep increasing. right? Can you add a few lines about what happens for general $SO(3)$? $\endgroup$
    – user6818
    Jul 29, 2012 at 16:27
  • $\begingroup$ @user6818: The word "increasing" is not the right term for the highest weights for $U(n)$ or $SU(n)$, they just are different each time. Here the $n-1$-st exterior power is in fact the dual representation of the defining one. $\endgroup$
    – Marc van Leeuwen
    Jul 29, 2012 at 18:27

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