Let $\mathbf x = (x_{i,j})_{1\leq i \leq n, 1\leq j \leq N}$ denote a collection of indeterminates. The algebraic group $\mathrm{SL}_n(\mathbb C)$ acts on $\mathbb C[\mathbf x]$ by "matrix multiplication", and invariant theory guarantees that the ring of invariants $\mathbb C[\mathbf x]^{\mathrm {SL}_n(\mathbb C)}$ is generated by certain "bracket quantities" $[i_1, \dots, i_n] = \det((x_{i,i_j})_{1\leq i,j\leq n})$, for $1\leq i_1 < \dots < i_n \leq N$.

(Edit: rewrote question; see Levent's comment) Is it true that $\mathrm{Frac}(\mathbb C[\mathbf x]^{\mathrm {SL}_n(\mathbb C)}) = \mathbb C(\mathbf x)^{\mathrm {SL}_n(\mathbb C)}$? In other words, can any rational function invariant under this action of $\mathrm{SL}_n$ be expressed as a quotient of two $\mathrm{SL}_n$-invariant polynomials?

  • 2
    $\begingroup$ Take any polynomial $f$ that is not an invariant. Then $f/f$ is invariant but $f$ is not so it cannot be written as a polynomial in the brackets. I think a better question would be "Does the fraction field of $\mathbb{C}[x]^{SL_n}$ equal $\mathbb{C}(x)^{SL_n}?$". $\endgroup$
    – Levent
    Sep 7, 2020 at 19:50
  • $\begingroup$ Thanks, this is indeed the question I meant to ask. $\endgroup$
    – Linus S
    Sep 7, 2020 at 20:03
  • $\begingroup$ Then let me answer. $\endgroup$
    – Levent
    Sep 7, 2020 at 20:04

1 Answer 1


As I mentioned in the comments, this is not true if we pick a non-invariant $f$ then $1=f/f$ is still an invariant. However we can ask the following: If $f/g$ is an invariant can we always find invariants $F,G$ such that $f/g=F/G$? In the case of an $SL_n$ action, this is true.

Say $f/g$ is an invariant, i.e. $$\frac{f}{g}=\frac{h\cdot f}{h\cdot g}$$ for all $h\in SL_n$. This is equivalent to the statement $(h\cdot g)f=(h\cdot f)g$ for all $h\in SL_n$. Without loss of generality, we may assume that $f$ and $g$ are coprime, i.e. $(f)\cap (g)=(fg)$ (the parentheses denote the ideal generated by the polynomial). Then $(h\cdot g)f$ is in $(fg)$ using the equality. As $h\cdot g$ has the same degree as $g$, we deduce that for all $h\in SL_n$, $h\cdot g=\lambda(h) g$ for some $\lambda(h)\in\mathbb{C}^{\times}$. Now, it is easy to show that $\lambda:SL_n\rightarrow\mathbb{C}^{\times}$ is necessarily a group homomorphism. But, there is no non-trivial group homomorphism $\lambda:SL_n\rightarrow\mathbb{C}^{\times}$! Hence, $g$ is an invariant. Similarly, $f$ is an invariant and the result follows.

Note that here we used the fact that there is no non-trivial group homomorphism $SL_n\rightarrow\mathbb{C}^{\times}$. This is not true for other groups, such as $GL_n$. In fact, the result does not hold for $GL_n$. Take the action of $GL_n$ on the vector space that you consider. Then, brackets are no longer invariants but any quotient $[i_1,i_2,\dots,i_n]/[j_1,j_2,\dots,j_n]$ of the brackets is an invariant.

  • $\begingroup$ Beautiful proof. Thank you! $\endgroup$
    – Linus S
    Sep 7, 2020 at 20:25
  • $\begingroup$ Thank you for the nice comment. $\endgroup$
    – Levent
    Sep 7, 2020 at 20:35

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