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I'm studying some representation theory on $S_n$ and $GL(V)$ and tensor spaces, and have come across a lot of material involving rational representations. I'm not really an algebraic geometer by trade, though I have covered some basics before, and I am essentially looking for some intuition behind what a rational map actually is, beyond the algebraic geometry definition. In my opinion, the (formal) definition is as ugly as sin, and any "informal" definition which I sometimes find accompanying the formal one is almost equally dire.

I feel like, given its definition as a "partial function between algebraic varieties", and also its name, it is in some manner an extension of the idea of a quotient of 2 polynomial functions $f(X)/g(X)$ (which would of course be defined everywhere where $g$ is nonzero). However, none of the materials I can find online are proving very fruitful in developing an understanding of what these maps are really meant to "be", beyond the raw definition.

I would just like to hear, in simple terms if possible, what you understand a rational map to be, if there is actually more to be said than just the definition: for example, is there some result which says "on nice (affine, irreducible, etc?) spaces, a rational map is of the form $f/g$, polynomials $f$, $g$" as above, or something along those lines? Or, are there just certain spaces (I normally only need to worry about affine things like $\mathbb{C}^n$, for example!), where a rational map always has some nice form? I should add, I don't particularly have any reason to think a rational map should be some quotient $f/g$ of polynomials, except for the terminology and the fact that the definition seems to permit such maps. If I am in totally the wrong ballpark, please let me know as well, I am simply seeking a better understanding of what the definition is really saying, even if there are cases where such maps always look nice and cases where they don't.

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The general notion of rational map from algebraic geometry is not particularly relevant to the notion of rational representations of linear algebraic groups (such as $GL_n$), which is what you actually seem to be interested in.

To say that a representation $\rho$ of $GL_n$ is rational is simply to say that there is a polynomial formula involving the matrix entries of $g \in GL_n$, together with perhaps the inverse of the determinant $\det(g)^{-1}$, which expresses $\rho(g)$ in terms of $g$.

Symmetric and exterior powers of the standard representation have this property, as does the determinant (thought of as a one-dimensional representation), and the tensor product of any two representations of this kind is again a representation.

The terminology rational representation is (in my view) somewhat outdated, and (in my experience) lots of people nowadays simply say algebraic representation instead (which I think also better captures the idea).


The notion of rational map in algebraic geometry is supposed to capture the idea of maps between varieties which are not every defined, due to possible denominators in the expressions that describe them. In rational representations, the only denominators which can appear are powers of the determinant, and these are everywhere defined on $GL_n$. So general discussions of rational maps between varieties won't be very helpful or relevant for your particular concern.

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    Ah - see, now why can't I ever find any sources which say that? My notes state that "a $GL(V)$-module is "rational" if there exist rational functions $f_{ij}(X_{rs}) \, (1 \leq r,\,s \leq \operatorname{dim}(V) = m,\,1 \leq i,\,j \leq \operatorname{dim}(M) = n)$ in $m^2$ variables $X_{rs}$ such that the composite map $GL_m(\mathbb{C}) \to GL(V) \to \operatorname{End}_\mathbb{C}(M) \to M_n(\mathbb{C}$, the first and third arrows obtained by taking bases and the middle arrow from the representation. I have just realised a rational function and a rational map may not be the same thing though...2012-03-27
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    ...Is a rational function just the ratio of 2 polynomial functions as I suggested? I think the wikipedia article may have just confused me after misunderstanding my notes.2012-03-27
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    @Spyam: Dear Spyam, A rational map is a map whose coordinates are rational functions, i.e. ratios of polynomials. But since your representation is defined at every point of $GL(V)$ (just because this is part of the definition of a representation), it turns out that the only demoninators that can appear are powers of $\det$. The "rational representation" terminology dates from a certain period in the development of the theory of algebraic groups, and reflects a certain viewpoint on the interaction between that theory and general concepts from algebraic geometry which I personally find ...2012-03-27
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    ...slightly misdirected. I have seen more than one professional representation-theorist become confused by the expression "rational representation"; as I already wrote, my experience is that most people (at least currently) prefer to say "algebraic representation". Regards,2012-03-27
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    Many thanks for your helpful responses on this. Just 2 quick further questions, if you don't mind. First of all, just to check, is my definition of a 'rational representation' in the above comment precisely equivalent to yours, that $\rho(g)$ is a polynomial in powers of $g$ with reciprocals of the determinant? (Similarly, polynomial and n-homogeneous representations are ones such that $\rho(g)$ is polynomial/n-homogeneous in $g$? The definitions I'm working with are slightly uglier and more unwieldy, but may plausibly be equivalent.)2012-03-28
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    Second question: how do we know that the only things in the denominator must be powers of det? Since we're working with invertible maps in GL the determinant must be nonzero sure, but why couldn't we have other "nonzero things" in the denominators? I was hoping you could elaborate on your comment "it turns out that the only denominators that can appear are powers of det"; isn't det just an element of our field ($\mathbb{C}$) anyway, so isn't "powers of det" a bit redundant for a polynomial equation? Or should our coefficients otherwise all be integers? Sorry if this is a stupid question.2012-03-28
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    @Spyam: Dear Spyam, Yes, if you have a representation $GL(V) \to GL(W)$ which is rational in the sense of your comment, then, after choosing bases, then the map $g \mapsto \rho(g)$ is necessarily given by a formula where each entry of the matrix $\rho(g)$ is given by some polynomial involving the entries of $g$ together with $\det(g)^{-1}$. This is perhaps not immediately obvious (i.e. it's not obvious that other, more complicated, rational functions aren't allowed), but if follows from basic facts in algebraic geometry. The key fact is that $GL(V)$ is actually an affine variety, and ...2012-03-28
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    ... its coordinate ring is generated by the matrix entries $x_{ij}$, together with $\det(g)^{-1}$. Regards,2012-03-28
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    @Spyam: Dear Spyam, one more thing: the key thing about "powers of $\det$" is that *negative* powers are allowed (and $\det(g)^{-1}$ is *not* a polynomial in the entries of $g$, but rather is a rational function, i.e. a ratio of polynomials; more precisely, $1$ is the numerator and the polynomial $\det(g)$, written out as a degree $n$ homogeneous expression in the $x_{ij}$s, is the denominator). Regards,2012-03-28
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    @Spyam: Just to fix ideas, suppose that our matrices are $2\times 2$, with entries denoted $a,b,c,d$ as usual. Then the contragredient of the standard representation is given by the formula $$\begin{pmatrix}a & b \\c & d \end{pmatrix} \mapsto \begin{pmatrix} d/(ad-bc) & -c/(ad-bc) \\ -b/(ad-bc) & a/(ad-bc) \end{pmatrix}.$$ The entries of the image are rational functions in the entries of the argument, and the only denominator that appears is $\det$, i.e. $ad - bc$. Regards,2012-03-28
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    That's very helpful, the example helps solidify the concepts: thanks again!2012-03-28