6 gave more readily accessible references
http://www.springerlink.com/content/wm3318667411003m/my article with Chipalkatti:"The bipartite Brill-Gordan locus and angular momentum". Transform. Groups 11 (to which 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).

this paperother 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

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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:

1. the orbit of $0$, with of course $S=T=0$.
2. the orbit of $x^4$ (all multiples are related by $G$), with $S=T=0$.
3. the orbit of $x^3 y$ (all multiples are related by $G$), with $S=T=0$.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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 http://www.springerlink.com/content/wm3318667411003m/

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

For case $4$ you can find defining equations in this paper.

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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 $C^5$. \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: 1. the orbit of$0$, with of course$S=T=0$. 2. the orbit of$x^4$(all multiples are related by$G$), with$S=T=0$. 3. the orbit of$x^3 y$(all multiples are related by$G$), with$S=T=0$. 4. 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. 5. orbits of$12\beta x^2 y (x+y)$,$\beta\neq 0$, with$(S,T)=(12\beta^2, -8\beta^3)$. Here different$\alpha$'s \beta$'s give different orbits.
6. 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.
7. 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.
8. 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 http://www.springerlink.com/content/wm3318667411003m/

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

For case $4$ you can find defining equations in this paper.

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