The ring of SL_2 invariants in sums of conjugation and tautological modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:22:28Z http://mathoverflow.net/feeds/question/59248 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59248/the-ring-of-sl-2-invariants-in-sums-of-conjugation-and-tautological-modules The ring of SL_2 invariants in sums of conjugation and tautological modules Greg Muller 2011-03-22T23:05:18Z 2011-08-07T02:22:12Z <h3>Rings of Invariants</h3> <p>Consider $G=SL_2(\mathbb{C})$, and let $V$ be a finite-dimensional $G$-representation. Let $\mathbb{C}[V]$ denote the ring of polynomial functions on the space $V$; it is a free polynomial ring in $\dim(V)$-many variables. Then the <strong>ring of invariants</strong> of $V$ is the subring $$\mathbb{C}[V]^G:=(f\in\mathbb{C}[V] s.t. \forall g\in G, f(gv)=f(v))$$</p> <h3>The Conjugation Representation</h3> <p>Let $M_2(\mathbb{C})$ denote the space of $2\times 2$ complex matrices. The <strong>conjugation</strong> representation $V_C$ of $G$ on $M_2(\mathbb{C})$ is defined by $g\cdot m:= gmg^{-1}$. Thus, $V_C$ is a 4-dimensional $G$-representation.</p> <p>The ring of invariants $\mathbb{C}[V_C]^G$ is well-known. It is freely generated by the functions $tr$ and $det$; this is a special case of the general fact that the only conjugation-invariant functions on $M_n(\mathbb{C})$ are symmetric functions of eigenvalues.</p> <h3>The Tautological Module</h3> <p>Let $V_T$ denote the $2$-dimensional $G$-representation where $G$ acts in the 'obvious' way, by the inclusion $G=SL_2(\mathbb{C})\subset GL_2(\mathbb{C})$. This is called the <strong>tautological</strong> $G$-representation. Since $SL_2(\mathbb{C})$ acts on $V_T$ with a dense orbit, the ring of invariants $\mathbb{C}[V_C]^G$ is boring; it is just $\mathbb{C}$.</p> <p>Now, let $V_T^n$ denote the $n$-fold direct sum of $V_T$. This is a natural $2n$-dimensional $G$-representation. We can choose a $G$-invariant skew-symmetric bilinear form $\omega$ on $V_T$; this is big words for the usual scalar cross product $v_1\times v_2$. This defines a natural $G$-invariant function on the space of pairs $(v_1,v_2)\in V_T^2$.</p> <p>Then the ring of invariants $\mathbb{C}[V_T^n]^G$ is generated by $\omega(v_i,v_j)$ for $1\leq i j\leq n$. However, these do not (in general) freely-generate the ring of invariants, there are relations between them. The relations are all of the form $$\omega(v_i,v_k)\omega(v_j,v_l)=\omega(v_i,v_j)\omega(v_k,v_l)+\omega(v_l,v_i)\omega(v_j,v_k)$$ for $1\leq ijkl\leq n$.</p> <p>Both of these facts can be deduced by observing that $\mathbb{C}[V_T^n]^G$ can be identified with the homogeneous coordinate ring of the Grassmanian $Gr(2,n)$. Then the generators above are the Plucker coordinates, and the relations are the 3-term Plucker relations (there are no higher Plucker relations here).</p> <h3>The Question</h3> <p>Both of these examples have clever solutions and pretty answers. I am curious about the combination of both cases. </p> <p>Let $m$ and $n$ be positive integers, and consider the direct sum $G$-representation $V^m_C\oplus V_T^n$. <strong>What is the ring of invariants $\mathbb{C}[V_C^m\oplus V_T^n]^G$?</strong></p> <p>I am aware of general procedures for producing these rings, for arbitrary finite-dimensional $G$-reps; see e.h. Sturmfel's <em>Algorithms in Invariant Theory</em>. However, I suspect that there is a clever solution to this particular problem. Not only because it is a combination of two problems with a clever solution, but because I have a guess as to what the answer is, and all the relations seem to be similar to the 3-term Plucker relations. I also suspect the answer is <em>ancient</em> (like much invariant theory), which is why I am trying to find an answer rather than try to prove my guess is correct by brute force.</p> http://mathoverflow.net/questions/59248/the-ring-of-sl-2-invariants-in-sums-of-conjugation-and-tautological-modules/71095#71095 Answer by David Wehlau for The ring of SL_2 invariants in sums of conjugation and tautological modules David Wehlau 2011-07-24T02:09:33Z 2011-07-24T02:09:33Z <p>I don't have the literature with me, but yes the answer to your question was well known to classical invariant theorists. My recollection is that it is one of the examples (sections) in Grace and Young (J.H. Grace and A. Young, The algebra of invariants, Cambridge Univ. Press, Cambridge, 1903). They give an answer without using polarization. All the generating invariants are the ones you describe or else invariants of degree 3 which are quadratic in (one or 2) of the tautological reps and linear in a conjugation rep. Describing the relations is harder but was understood classically. I am not sure of a reference though. </p>