Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$ - MathOverflow

Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$

2010-12-15T21:56:32Z

One sees that given the $SL(2,\mathbb{C})$ action on $\mathbb{C}^5$, thought of as the space of polynomials of the form,

$$a_0 x^4 + 4a_1 x^3 y + 6a_2x^2y^2 + 4a_3xy^3 + a_4 y^4$$

the ring of invariants is generated by the following functions,

$$g_2(a) = a_0a_4 - 4a_1 a_3 + 3a_2^2$$

and

$$g_3(a) = a_0a_2a_4 - a_0a_3^2 - a_1^2a_4 + 2a_1a_2a_3 - a_2^3$$

But if these $g_2$ and $g_3$ satisfy the discriminant $=0$ condition then there are inequivalent $SL(2,\mathbb{C})$ polynomials which map to the same $(g_2,g_3)$ point.

But if I look at say Theorem 5.9 in the book by Mukai then I get to see that the closure equivalent classes of orbits of the action of a linearly reductive group like $SL(2,\mathbb{C})$ on $\mathbb{C}^5$ are in bijective correspondence to the the points of $\mathbb{C}^5//SL(2,\mathbb{C})$ (which is defined as the spectrum of the invariant polynomials under $SL(2,\mathbb{C})$)

Also look at the theorem at the end of page 11 of this paper.

In the above paper "//" is defined as identifying points in the affine variety if one lies in the closure of the orbit through the other.

Are these two notions of "//" equivalent? If yes, how?
In the light of the above two theorems, can one say that the $SU(2)$ invariant polynomials among binary homogeneous quartics are in bijection with those closure equivalent classes of orbits of $SU(2)^{\mathbb{C}} = SL(2,\mathbb{C})$ which are labeled by the pairs of invariants $(g_2,g_3)$ such that $g_2^3 - 27 g_3^2 \neq 0$ ?
Hence if I am interested in only closure equivalent orbits can I just forget those pairs of values of the invariants which lie on the discriminant $0$ curve?
Conversely given a $(g_2,g_3)$ for which the above discriminant condition is satisfied can one write down the family of $SU(2)$ invariant polynomials explicitly?
I would anyway like to know how to distinguish the orbits corresponding to the discriminant $0$ condition.

In light of the various extremely helpful and references that have come up, I realize that there there is a notion of a "discriminant" for homogeneous polynomials of degree $d$ in $n$ variables. (call this space of polynomials as $P(n,d)$) This discriminant is in some sense a "homogeneous invariant" and for the $n=2$ case in which I am interested in, it generates the subalgebra of the coordinate ring over of these polynomials which is invariant under this group action (call that $\mathbb{C}[P(n=2,d=4)]^{SL(2,\mathbb{C})}$). (this is lucky!)

I guess the discriminant in this case has to be a polynomial in polynomials of homogeneous degree $4$ in $2$ variables. The above I guess implies that $\mathbb{C}[P(n=2,d=4)]^{SL(2,\mathbb{C})}$ is generated by just one such polynomial in polynomials.

I would like to know how is a ``discriminant" defined for such polynomials. (searching and asking around I am only getting definitions for the single variable case)

I want to know if knowing the generator of $\mathbb{C}[P(n=2,d=4)]^{SL(2,\mathbb{C})}$ tells me about the initial objective of knowing $P(n=2,d=4)\text{ }mod\text{ }SL(2,\mathbb{C})$

I wonder if this unique generator of the invariant subalgebra is related to the null-cone that was pointed out by Bart in his comment.

Also I would like to be pointed out if there is any mistake in what I said above!

I had recently tried asking a similar question here. But I think I could not precisely convey what I was looking for. Let me here try to give a specific situation that I need to understand coming from certain other considerations in Superconformal Quantum Field Theories.

I can think of $\mathbb{C}^5$ as being the space of all homogeneous degree $4$ polynomials in $2$ variables. On this space $SL(2,\mathbb{C})$ has the standard action.

I want to know what is the most explicit (or the best!) way to describe the quotient space thus obtained. I want to understand how do the orbits look like.

[EDIT: I was initially asking if there exists fixed subspaces etc but then from the comments I realized that I was missing the elementary fact that it is an irreducible representation! Hence nothing like this can exist.]

I tried something naive. I wrote down the most general element of $SL(2,\mathbb{C})$ using its canonical polar decomposition and then acted it on the most general homogeneous degree $4$ polynomial in $2$ variables and tried to see how the coefficients change. Unfortunately the equations are very complicated and I didn't see any hope of me being able to solve them to find the fixed points.

Apart from this specific example I would also like to know of references to simpler examples than this where a similar question is asked and answered.

Answer by robot for Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$

2010-12-16T00:38:15Z

I think that for irreducible (linear) representations of algebraic groups there is a classification of orbits in terms of Bruhat ordering of the Weyl group. Unfortunately, I don't recall a reference. But see Remark 3.2.8 in Parabolic Geometries.

Answer by Bart Van Steirteghem for Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$

2010-12-16T03:28:10Z

The general problem of describing the orbits in a finite dimensional representation of $SL(2,\mathbb{C})$ seems (I have no access to the paper) to have been solved by V.L.Popov in "Structure of the closure of orbits in spaces of finite-dimensional linear SL(2) representations", [Matematicheskie Zametki, Vol. 16, No. 6, pp. 943–950, December, 1974].

To get a glimpse of Popov's theorem, you can look at the Zentralblatt review: http://www.zentralblatt-math.org/zmath/scans.html?volume_=313&count_=115

For the case of irreducible representations, Popov refers to a 1966 paper by Hadžiev, Dž: [Dokl. Akad. Nauk UzSSR 1966 no. 12, 3–6.]

The first step in describing the $SL(2,\mathbb{C})$-orbits in the space $R_n=\mathbb{C}[X,Y]_n$ of binary forms of degree $n$ ($n=4$ in the original question) is the following lemma (see e.g., p.30 in [H.P. Kraft, Geometrische Methoden in der Invariantentheorie]): a) every binary form $f \in R_n$ is the product of linear forms; b) for every nonzero binary form $f \in R_n$ there exist integers $r \ge s \ge 0$ and a binary form $f' \in R_{n-r-s}$ which has no linear factor of multiplicity $>s$ and which is not a multiple of $X$ or $Y$, such that $f$ lies in the orbit of $X^rY^sf'$

Answer by Mikhail Borovoi for Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$

2010-12-17T10:04:54Z

See V.L. Popov, E.B. Vinberg, Invariant Theory, Encyclopaedia of Mathematical Sciences, Vol. 55, Algebraic Geometry IV, Springer-Verlag, Berlin, 1994.

You can find the invariants of a binary quartic in 0.12 (p. 141). Fixing values of the two basic invariants, you obtain a closed set, which contains only finitely many orbits (Theorem 8.8, p.247) and only one closed orbit (Corollary of Theorem 4.7, p.189). Within such a closed set, the classification is given by means of multiplicities of roots, as in Popov's note cited by Bart.

Answer by Abdelmalek Abdesselam for Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$

2010-12-22T16:27:12Z

The description of orbits in the $d=4$ and $n=2$ is not difficult.

It is better to work projectively i.e. look at the orbits in the projective space $P(\mathbb{C}^5)$ of nonzero binary quartics up to a constant. A nonzero binary quartic $F$ can be written as a product of linear forms and therefore corresponds to a collection of four points on the projective line with possible repetitions but no ordering. If all points are distinct one can transform them by an $SL_2$ element into 0, 1, $\infty$ and some other guy $\lambda$. Orbits in this case are in one-to-one correspondence with the $j$-invariant $$j(F)=\frac{4}{27}\times\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}$$ Essentially $\lambda$ is the cross ratio of the four points but this depends on the chosen ordering. The above rational fraction is what is needed in order to kill the dependence on the ordering. The same $j$-invariant can be expressed in terms of invariants of $F$: $$ j(F)=\frac{S^3}{S^3-27 T^2} $$ where $S$ is the invariant of degree 2 and $T$ is the invariant of degree 3 (both defined up to normalization by a constant). They generate the ring of invariants you are talking about. What you said above is false: this ring is not generated by the discriminant which is not a fundamental invariant. It is given by the denominator $S^3-27 T^2$. Finally the other orbits correspond to three points or less. These points can be placed anywhere we want on the projective line by an $SL_2$ element. So these orbits are characterized by the multiplicities $(2,1,1)$, $(2,2)$, $(3,1)$ and $(4)$. The last two form the null cone of binary forms with a root of multiplicity $>$ half the degree of the form. All invariants vanish on these ones. What one needs to distinguish them are covariants: joint invariants of the form and an extra auxiliary point. For $n>2$ one needs mixed concomitants, i.e., joint invariants of the form and an extra auxiliary complete flag.

What I said above is a description of the quotient $Q=P(\mathbb{C}^5)/G$ where $G=SL_2$. Now if you want the affine version $A=\mathbb{C}^5/G$ it can be described as follows. If $F$ is a nonzero binary quartic denote by $[F]$ the corresponding point in projective space, i.e., its class up to multiplication by a nonzero constant. On $\mathbb{C}^5$ one has the composition of maps $F\rightarrow [F] \rightarrow G[F]$ which ends in $Q$. Clearly the preimage of each point in $Q$ is a union of orbits in $\mathbb{C}^5$. Figuring out $A$ from the previous description of $Q$ can be done by looking at each $G[F]$ and asking if different multiples of that $F$ are related by $G$. The resulting list of affine orbits is:

the orbit of $0$, with of course $S=T=0$.
the orbit of $x^4$ (all multiples are related by $G$), with $S=T=0$.
the orbit of $x^3 y$ (all multiples are related by $G$), with $S=T=0$.
orbits of $6\alpha x^2 y^2$, $\alpha\neq 0$, with $(S,T)=(3\alpha^2, -\alpha^3)$. Here different $\alpha$'s give different orbits.
orbits of $12\beta x^2 y (x+y)$, $\beta\neq 0$, with $(S,T)=(12\beta^2, -8\beta^3)$. Here different $\beta$'s give different orbits.
orbits of $\gamma xy(x-y)(x-\lambda y)$ with $\gamma\neq 0$, and $\lambda$ in a fundamental domain of the complex plane minus the points $0,1,-1,2,\frac{1}{2}, -\omega, -\omega^2$, where $\omega=e^{\frac{2i\pi}{3}}$, with respect to the group ($\simeq S_3$) of six transformations generated by $\lambda\rightarrow 1-\lambda$ and $\lambda\rightarrow \frac{1}{\lambda}$. One then has $$ (S,T)=\left(\frac{\gamma^2}{12} (\lambda^2-\lambda+1) ,\frac{\gamma^3}{2^4\times 3^3}(\lambda+1)(2\lambda^2-5\lambda+2)\right) $$ Again different pairs $(\gamma,\lambda)$ give different orbits.
orbits of $\mu xy(x-y)(x+\omega y)$, with $\mu$ in a fundamental domain of $\mathbb{C}\backslash{0}$ with respect to multiplication by a cube root of unity. One then has $$ (S,T)=(0,\frac{i\sqrt{3}}{2^4\times 3^2} \mu^3) $$ One needs this fundamental domain to avoid repetition because multiplying such a form by $\omega$ does give an $SL_2$ equivalent form, and this only happens when one multiplies by a cube root of unity.
orbits of $\nu xy(x-y)(x+y)$, with $\nu$ in a fundamental domain of $\mathbb{C}\backslash{0}$ with respect to multiplication by $-1$. One has $(S,T)=(\frac{\nu^2}{4},0)$. Same remark that minus such a form is $SL_2$ equivalent to the original one.

Now if you look at the map from $A$ to $\mathbb{C}^2$ given by the pair $(S,T)$ you see the following. Each pair away from the $S^3-27 T^2=0$ discriminant curve has exactly one preimage. This corresponds to the cases $6,7,8$ above. The last two are called equianharmonic and harmonic configurations respectively. Then if $(S,T)\neq (0,0)$ is on the discriminant curve, one has exactly two preimages, cases $4,5$ with $\alpha=2\beta$. Finally the point $(0,0)$ has three preimages, the cases $1,2,3$.

If you want to distinguish between the first five cases invariants will not do, you need covariants. These cases are examples of so called coincident root loci, you can lookup the papers by my collaborator J. Chipalkatti for the equations for such things. See also http://mathoverflow.net/questions/13672/polynomial-with-two-repeated-roots

Case $2$ is characterized by the vanishing of the Hessian of $F$, and the inequality $F\neq 0$. You can find equations for case $3$ in my article with Chipalkatti: "The bipartite Brill-Gordan locus and angular momentum". Transform. Groups 11 (2006), no. 3, 341--370. A preprint version is http://arxiv.org/abs/math.AG/0502542

(you have to add the inequality that the Hessian is not zero).

For case $4$ you can find defining equations in this other article with Chipalkatti: "Brill-Gordan Loci, transvectants and an analogue of the Foulkes conjecture". Adv. Math. 208 (2007), no. 2, 491--520. A preprint version is http://arxiv.org/abs/math.AG/0411110