# Is an SL-invariant rational function necessarily a quotient of two SL-invariant polynomials?

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?

• 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 $\mathbb{C}(x)^{SL_n}?$". Sep 7, 2020 at 19:50
• Thanks, this is indeed the question I meant to ask. Sep 7, 2020 at 20:03
• Then let me answer. Sep 7, 2020 at 20:04

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.