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$\newcommand\Alt{\bigwedge\nolimits}$Let $G=\operatorname{SL}(2,\Bbb C)$, and let $R$ denote the natural 2-dimensional representation of $G$ in ${\Bbb C}^2$. For an integer $p\ge 0$, write $R_p=S^p R$; then $R_1=R$ and $\dim R_p=p+1$.

Using Table 5 in the book of Onishchik and Vinberg, I computed that the representation $$ R_2\otimes\Alt^2 R_4 $$ contains the trivial representation with multiplicity one. I used the table as a black box.

Question. Let $V\subset R_2\otimes\Alt^2 R_4$ denote the corresponding one-dimensional subspace. How can one describe $V$ as a subspace geometrically?

Motivation: I want to consider a $\operatorname{PGL}(2,k)$-fixed trivector $$v\in V\subset R_2\otimes\Alt^2 R_4\subset \Alt^3(R_2\oplus R_4)$$ of the 8-dimensional vector space $W=R_2\oplus R_4$ over a field $k$ of characteristic 0, and then to twist all this using a Galois-cocycle of $\operatorname{PGL}(2,k)$. For this end I need a geometric description of $V$.

Feel free to add/edit tags!

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    $\begingroup$ @LSpice as the representation is algebraic, yes it's clear since $k$ is infinite and hence $\mathrm{SL}(2,k)$ is Zariski-dense. $\endgroup$ – YCor Nov 24 '20 at 17:38
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    $\begingroup$ @LSpice ah I assumed it's a subfield of $\mathbf{C}$, but anyway Mikhail will eventually clarify. $\endgroup$ – YCor Nov 24 '20 at 17:45
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    $\begingroup$ Seeing 8 and 3 here makes me wonder whether there's some way to find a $G = \operatorname{SL}_2$ inside $\tilde G = \operatorname{Spin}_8$ so that the spinor representation of $\tilde G$ becomes the $R_2 \oplus R_4$ representation of $G$, and then use triality somehow. $\endgroup$ – LSpice Nov 24 '20 at 18:07
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    $\begingroup$ @LSpice: There is a construction of octonions as pairs $(s,S)$ where $s$ is a scalar and $S$ is a binary sextic. Multiplication is done in an $SL_2$ invariant way using the third transvectant (to produce another sextic) and the sixth transvectant (to produce a scalar). See this article by Dixmier degruyter.com/view/journals/crll/1984/346/article-p110.xml $\endgroup$ – Abdelmalek Abdesselam Nov 24 '20 at 19:29
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    $\begingroup$ Not sure about if $G_2$ helps for the OP's question. But as far a connection between $G_2$ and $SL_2$, Dixmier's article is definitely relevant. $\endgroup$ – Abdelmalek Abdesselam Nov 24 '20 at 19:49
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Here's another very nice (but still algebraic) interpretation that explains some of the geometry: Recall that $\operatorname{SL}(2,\mathbb{C})$ has a $2$-to-$1$ representation into $\operatorname{SL}(3,\mathbb{C})$ so that the Lie algebra splits as $$ {\frak{sl}}(3,\mathbb{C}) = {\frak{sl}}(2,\mathbb{C})\oplus {\frak{m}} $$ where ${\frak{m}}$ is the ($5$-dimensional) orthogonal complement of ${\frak{sl}}(2,\mathbb{C})$ using the Killing form of ${\frak{sl}}(3,\mathbb{C})$. Note that ${\frak{m}}$ is an irreducible ${\frak{sl}}(2,\mathbb{C})$-module, and that every element $x\in {\frak{sl}}(3,\mathbb{C})$ can be written uniquely as $x = x_0 + x_1$ with $x_0\in {\frak{sl}}(2,\mathbb{C})$ and $x_1\in{\frak{m}}$. Note also that $[{\frak{m}},{\frak{m}}]= {\frak{sl}}(2,\mathbb{C})$.

This defines the desired pairing ${\frak{sl}}(2,\mathbb{C})\times \bigwedge\nolimits^2({\frak{m}})\to\mathbb{C}$: Send $(x_0,y_1,z_1)$ to $\operatorname{tr}(x_0[y_1,z_1])$. Of course, this makes the $\operatorname{SL}(2,\mathbb{C})$-invariance of the pairing obvious.

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  • $\begingroup$ The 2-to-1 representation is just the adjoint representation, right? $\endgroup$ – LSpice Nov 24 '20 at 20:01
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    $\begingroup$ @LSpice: Yes, of course. There's really only one (up to conjugacy). It's a little bit easier to see how everything works if you represent the image as ${\frak{so}}(3,\mathbb{C})\subset{\frak{sl}}(3,\mathbb{C})$, for then ${\frak{so}}(3,\mathbb{C})$ is the skew-symmetric matrices while $\frak{m}$ is the traceless symmetric matrices. This makes it clear that $[{\frak{m}},{\frak{m}}] = \mathfrak{so}(3,\mathbb{C})$ and shows exactly how everything works as an $\mathfrak{so}(3,\mathbb{C})$ representation. $\endgroup$ – Robert Bryant Nov 24 '20 at 21:18
  • $\begingroup$ Excellent! This is exactly what I need! $\endgroup$ – Mikhail Borovoi Nov 25 '20 at 8:06
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For a purely geometric construction, see further below, after the following algebraic considerations.

There is a Wronskian isomorphism which as a particular case says that the second exterior power of $R_4$ is isometric to the second symmetric power of $R_3$. So the invariant in question is $I(Q,C)$, a joint invariant in a binary quadratic $Q$ and a binary cubic $C$, which is linear in $Q$ and quadratic in $C$. This is indeed unique up to scale and is given in classical symbolic notation (see, e.g., Grace and Young) by $$ (ab)(ac)(bc)^2 $$ where $Q=a_{x}^{2}$ and $C=b_{x}^{3}=c_{x}^{3}$.

Another construction is to start from the binary discriminant, and polarize it to get a bilinear form (the unique invariant one on $R_2$), and apply this bilinear form to $Q$ and the Hessian of $C$.

If one does not want to use the Wronskian isomorphism then the invariant would be $J(Q,F_1,F_2)$, trilinear in the quadratic $Q$ and the two binary quartics $F_1,F_2$. It would satisy the antisymmetry $J(Q,F_2,F_1)=-J(Q,F_1,F_2)$ and would be given in symbolic form by $$ (ab)(ac)(bc)^3 $$ where now $Q=a_{x}^{2}$, $F_1=b_{x}^{4}$, and $F_2=c_{x}^{4}$.


Geometric construction:

Consider $\mathbb{P}^1$ embedded by Veronese as a conic $\mathscr{C}$ in $\mathbb{P}^2$. A binary quadratic $Q$ corresponds to a point in $\mathbb{P}^2$. A binary cubic $C$ corresponds to a divisor or an unordered collection of three points $\{P_1,P_2,P_3\}$ on $\mathscr{C}$. Let $T_1, T_2, T_3$ be the tangents to the conic at $P_1,P_2,P_3$. Consider the points of intersection $T_1\cap P_2P_3$, $T_2\cap P_1P_3$, $T_3\cap P_1P_2$. They are aligned and thus define a line $L$. The vanishing of the invariant $I(Q,C)$ detects the situation where the point $Q$ is on the line $L$. I don't remember if the collinearity result I mentioned has a name, but it is a degenerate case of Pascal's Theorem.

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  • $\begingroup$ Thank you! I will try to read Grace and Young, the book published in 1903.... $\endgroup$ – Mikhail Borovoi Nov 24 '20 at 19:29
  • $\begingroup$ No pbm. For more geometry, it's probably possible to express the vanishing of $I(Q,C)$ as a special configuration of three points on a conic (the roots of the cubic) in the projective plane and a point in the plane corresponding to Q. $\endgroup$ – Abdelmalek Abdesselam Nov 24 '20 at 19:47
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    $\begingroup$ I had never heard of Wronskian isomorphisms, and the first Google result was another answer of yours. Are the papers by you and Chipalkatti, On Hilbert covariants and On the Wronskian combinants of binary forms, referenced there the best place to learn about this? $\endgroup$ – LSpice Nov 24 '20 at 20:09
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    $\begingroup$ Of course....;) $\endgroup$ – Abdelmalek Abdesselam Nov 24 '20 at 20:09

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