GL(V)-representation theory for a Lie bracket kernel - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:31:46Z http://mathoverflow.net/feeds/question/65901 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65901/glv-representation-theory-for-a-lie-bracket-kernel GL(V)-representation theory for a Lie bracket kernel Jim Conant 2011-05-24T21:09:48Z 2011-09-20T14:50:49Z <p>Let $V$ be a vector space over a field of characteristic $0$, and let $L_k(V)$ be the degree $k$ part of the free Lie algebra over $V$. There is an exact sequence $$0\to D_n(V)\to L_1(V)\otimes L_{n+1}(V)\to L_{n+2}V\to 0$$ where the map on the right is the Lie bracket and $D_n(V)$ is defined as the kernel of this map. Now $D_n(V)$ is a $GL(V)$-module, and I'm curious about how it decomposes as a direct sum of irreducibles. By playing around with dimension formulas, I was able to show that $D_1(V)=S^{(1,1,1)}V$, $D_2(V)=S^{(2,2)}V$ and $D_3(V)=S^{(3,1,1)}V$. Here $S^\lambda(V)$ is the irreducible representation of $GL(V)$ corresponding to partition $\lambda$. So in fact, if I haven't made a mistake, $D_k(V)$ is actually irreducible for $k=1,2,3$, with nice symmetric looking partitions.</p> <p>Is $D_k(V)$ always irreducible, and is there an easy way to tell what the partition is? If not, does anyone know what is known about this?</p> <p>If it helps, here is an alternate way of characterizing $D_n$. Consider the Lie operad, and look at the part with a total of $n+2$ input/output slots, denoted $Lie((n+2))$. Then $$D_n\cong [V^{\otimes n+2}\otimes Lie((n+2))]_{Sym(n+2)},$$ which denotes the coinvariants under the action of $Sym(n)$ acting simultaneously on $Lie((n+2))$ and $V^{\otimes n}$, in the latter case by permuting the factors. This is isomorphic to a space of planar unitrivalent trees with leaves labeled by vectors from $V$, modulo antisymmetry and Jacobi (IHX) relations. </p> <p>I don't know very much representation theory, so it's possible my question has an easy answer.</p> <p><strong>Added 9/20/2011</strong> <a href="http://arxiv.org/abs/1109.3963" rel="nofollow">Morita, Sakasai and Suzuki</a> just posted a preprint proving that $D_{4k+2}$ and $D_{4k+3}$ always have <em>symmetric</em> decompositions, in the sense that there is a symmetry in the corresponding Young diagrams exchanging rows and columns. This is a neat result.</p> http://mathoverflow.net/questions/65901/glv-representation-theory-for-a-lie-bracket-kernel/65946#65946 Answer by James Griffin for GL(V)-representation theory for a Lie bracket kernel James Griffin 2011-05-25T10:29:23Z 2011-05-25T10:50:43Z <p>The representations are not in general irreducible. The second characterisation is the more useful to my mind because one can decompose $Lie((n+2))$ as an $Sym(n+2)$-module, which gives a decomposition of the $GL(V)$-module of interest. Theo already recited the magical words Schur-Weyl.</p> <p>The dimension of $Lie((n+2))$ as a vector space is $n!$. So the $Sym(6)$-module $Lie((6))$ has dimension 24. However the representation of maximal dimension is given by the partition $(3,2,1)$ which has dimension 16. Playing with the dimensions along with a little knowledge of the Lie operad (that it doesn't have any 1 dimensional modules) gives the decomposition of dimension 24 = 10 + 9 + 5. The 10 and 9 dimensional modules are unique up to transposition. For the 5 dimensional one there are (up to transposition) two possibilities (2,2,2) and (2,1,1,1,1), but we know that it must be the first because restricting (2,1,1,1,1) to $Sym(5)$ gives a one dimensional module in $Lie(5)$.</p> <p>I think that I've read about the general case somewhere but my preliminary google has drawn a blank. If I can find the reference I'll add it to my answer.</p> <h2>Update</h2> <p>The $Sym(n+2)$-module $Lie((n+2))$ is known as the Whitehouse module, there are some slides on Richard Stanley's <a href="http://www-math.mit.edu/~rstan/trans.html" rel="nofollow">website</a>.</p> http://mathoverflow.net/questions/65901/glv-representation-theory-for-a-lie-bracket-kernel/67406#67406 Answer by Naoya Enomoto for GL(V)-representation theory for a Lie bracket kernel Naoya Enomoto 2011-06-10T05:27:40Z 2011-06-10T05:27:40Z <p>Hello. I would like to explain an elementary way to decompose $D_n(V)$ as a $GL(V)$-module. (But for large $n$, it is difficult to calculate by hand.) </p> <p>We will use the following results. From now, we assume $\dim{V} \gg n$. </p> <ol> <li><p>The multiplicities of an irreducible $GL(V)$-module $S^\lambda{V}$ in $L_n(V)$ is equal to the one of an the irreducible $\mathrm{Cyc}_n$-module $\exp(2\pi{i}/n)$ in the $S_n$-module $\mathrm{Res}_{\mathrm{Cyc}_n}^{S_n}D^\lambda$. Here $D^\lambda$ is the irreducible $S_n$-module corresponding to $\lambda$.</p></li> <li><p>There is a combinatorial way to decompose $\mathrm{Res}_{\mathrm{Cyc}_n}^{S_n}D^\lambda$. We use the notion of the "major index" for a standard tableau of shape $\lambda$. This appears in Stanley's slide which is referd by James Griffin. For more details about 1 and 2, see Garsia's paper "Combinatorics of the free Lie algebra and the symmetric group"(Theorem 8.4) and Reutenauer's book "Free Lie algebras"(Theorem 8.8 and 8.9) etc. </p></li> <li><p>Pieri's rule. We can obtain the irreducible decomposition of the $GL(V)$-module $L_1(V) \otimes S^\lambda{V}$, namely this module is isomorphic to the direct sum of irreducible $GL(V)$-modules $S^\mu{V}$, where $\mu$ runs over the set of partitions by adding a box to $\lambda$. Then, if we know the irreducible decomposition of $L_n(V)$ by 1 (and 2), we can also obtain the irreducible decomposition of $L_1(V) \otimes L_n(V)$. </p></li> </ol> <p>Example. For simplicity, we denote $\lambda$ by $S^\lambda{V}$.</p> <p>$L_5(V) \cong (4,1) \oplus (3,2) \oplus (3,1^2) \oplus (2^2,1) \oplus (2,1^3)$. </p> <p>$L_1(V) \otimes L_5(V)$ is isomorphic to $(5,1) \oplus 2(4,2) \oplus 2(4,1^2) \oplus (3^2) \oplus 3(3,2,1) \oplus 2(3,1^3) \oplus (2^3) \oplus 2(2^2,1^2) \oplus (2,1^4)$.</p> <p>$L_6(V) \cong (5,1) \oplus (4,2) \oplus 2(4,1^2) \oplus (3^2) \oplus 3(3,2,1) \oplus (3,1^3) \oplus 2(2^2,1^2) \oplus (2,1^4)$. Then we have $D_4(V) \cong (4,2) \oplus (3,1^3) \oplus (2^3)$.</p> <p>Similarly, I think, we can obtain $D_5(V) \cong (5,1^2) \oplus (4,2,1) \oplus (3^2,1) \oplus (3,2,1^2) \oplus (2^2,1^3)$.</p>