TITLE: Restriction of representation from GL(n) to O(n)
QUESTION [5 upvotes]: So my question is somewhat similar to Restriction from $\mathfrak{gl}_{2n}$ to $\mathfrak{sp}_{2n}$; but I was having difficulty understanding the formula given in reference (Harris & Fulton) mentioned there. 
Equation (25.37) in Harris & Fulton's "Representation Theory: A first course", says that if $m=2n$ or $m=2n+1$, $\lambda = (\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_m)$ is a $GL_m$ highest weight, and $\bar{\lambda}=(\bar{\lambda_1} \geq \cdots \geq \bar{\lambda}_n)$ is an $O_m$ highest weight, then the multiplicity of the irreducible highest-weight $O_m$ representation $V_{\bar{\lambda}}$ in the restriction of the irreducible highest-weight $GL_m$ representation $V_{\lambda}$ is equal to $\sum_{\eta} N_{\eta, \bar{\lambda}, \lambda}$; where the sum is over all partitions $\eta$ with all parts even, and $N_{\lambda_1, \lambda_2, \lambda_3}$ is the Littlewood-Richardson coefficient. I had some trouble getting this formula to work for $m=2$:
Q1: Since $O_m$ is disconnected, an irreducible $O_m$ representation is given by (a) if $m=2n+1$, a highest weight ${\lambda}=(\bar{\lambda_1} \geq \cdots \geq \bar{\lambda}_n)$ and a choice of $+$ or $-$ (b) if $m=2n$, a highest weight ${\lambda}=(\bar{\lambda_1} \geq \cdots \geq \bar{\lambda}_n)$ and, if $\bar{\lambda}_n=0$, a choice of $+$ or $-$. [See pg 53 of http://dspace.mit.edu/handle/1721.1/8642?show=fullfor more details.] In the above formula, the book seems to be considering $\mathfrak{o}_m$ representations instead, so how should the choices of $+$ or $-$ that arise be incorporated into the above formula?
Q2: Now take $m=2$. First let $\lambda=(1,1)$. If $\bar{\lambda} \neq 0$, then clearly $\sum_{\eta} N_{\eta, \bar{\lambda}, \lambda}=0$; if $\bar{\lambda}=0$, then $\sum_{\eta} N_{\eta, \bar{\lambda}, \lambda}=0$ also (since $N_{2,0,(1,1)}=0$). So this seems to be saying that the restriction of $V_{(1,1)}$ to $O(2)$ has no constituents, which is clearly false. The same problem occurs for $\lambda=(2k+1, 2k+1)$. Another example that was troubling me is $\lambda=(6,4), \bar{\lambda}=(4)$: then $N_{6, 4, (6,4)}=N_{(4,2), 4, (6,4)}=1$, so this says that the multiplicity of $V_{4}$ in the restriction of $V_{(6,4)}$ is $2$, which is also false.

REPLY [4 votes]: For the exterior powers (highest weight $(1,1,\dots,1)$), the restriction to the orthogonal group is still irreducible -- I think you are forgetting to take $\overline{\lambda} = 0$. 
There are also issues with stable ranges in this formula (you have to assume that $\lambda$ doesn't have more than $n$ parts). If you want the formulas to work in all cases, then you have to use modification rules for transforming partitions of length more than $n$ into those with length at most $n$ (sometimes you have to introduce signs to cancel terms). Details of this modification rule can be found in several sources. One such that I like is Koike--Terada's paper: http://www.sciencedirect.com/science/article/pii/0021869387900998 (Section 2.4)