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Let $X$ and $Y$ be two smooth (probably projective) algebraic varieties defined over $\mathbf{C}$. What is known in general about the (topological) space of holomorphic maps $\mathrm{Hol}(X(\mathbf{C}),Y(\mathbf{C}))$ from $X(\mathbf{C})$ to $Y(\mathbf{C})$?

In particular, I would be interested to know under what hypothesis on $X$ and $Y$ the space $\mathrm{Hol}(X(\mathbf{C}),Y(\mathbf{C}))$ is a smooth manifold (and to have a formula to compute the dimension of its connected components).

One example I have in mind is : if $X=Y=\mathbf{P}^1$ (the projective line) then $\mathrm{Hol}(\mathbf{P}^1(\mathbf{C}),\mathbf{P}^1(\mathbf{C}))$, in which case one gets the space of complex rational functions (which has connected components indexed by the positive integers (the degree), but each is a smooth algebraic variety).

I think this might be an (easy?) application of the theory of Grothendieck $\mathrm{Hom}$-schemes, but I don't feel very at ease with this. I would also be interested to know when $\mathrm{Hom}(X,Y)$ is a smooth algebraic variety.

Many thanks,


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What topology are you equipping the space Hol(X(C),Y(C)) with? – Harry Gindi May 31 '10 at 11:00
The topology of the subspace of the whole mapping space, equipped with the compact open topology. – Oblomov May 31 '10 at 11:29
By the way, if the answer to my question were true for an other topology, then please let me know too! – Oblomov May 31 '10 at 11:37
Assume $X$, $Y$ projective. Justifying example with proj. line (when corrected with constant maps) uses ${\rm{Pic}}(\mathbf{P}^1_ A) = \mathbf{Z}$ for local $A$. Nice exercise to prove Grothendieck's Hom-scheme equips set of holomorphic maps with compact-open topology. Hom-scheme is (by construction) countable disjoint union of quasi-proj. schemes; identifying when two maps lie on same connected or even irreducible component looks hopeless. The tangent spaces can be described (using tangent bundle of $Y$), but there's no clean "formula" for their dimensions. Smoothness should be rare. – BCnrd May 31 '10 at 13:57
@Oblomov: I read "positive degree" in English rather than French sense ($> 0$, not $\ge 0$), so sounds like ignoring deg. 0 maps, which are in Hom-scheme. To show Hom-scheme has expected topology on $\mathbf{C}$-valued points, tedious to fit argument in comment box. Here's a hint: use valuative criterion for properness on each component to prove that a projective embedding on $Y$ induces a closed immersion on Hom-schemes, so reduce to case $Y = \mathbf{P}^n$. Then study line bundles on $X$, univ. property of $\mathbf{P}^n$, & topological meaning of pts of id. component of Pic scheme of $X$. – BCnrd May 31 '10 at 16:18

2 Answers 2

up vote 7 down vote accepted

Let $X,Y$ be analytic spaces with $X$ compact reduced and $Y$ arbitrary (i.e. maybe with nilpotents in its structure sheaf). Then Douady showed in his thesis that the set of holomorphic maps $Hol(X,Y)$ can be endowed with the structure of an analytic space whose underlying topology is the compact-open topology. If $X,Y$ are compact manifolds, the Zariski tangent space at $f:X\to Y$ is a subspace of the finite-dimensional vector space of sections of the tangent bundle to $Y$ pulled-back to $X$ viz. $T_f(Hol(X,Y))\subset \Gamma(X,f^\star TY)$. I don't know any good general criterion for $Hol(X,Y)$ to be smooth at $f$.

Edit Here is a class of examples which might interest you, where smoothness occurs. Let $X$ be a Riemann surface of genus $g$. The space of ramified covers $f:X\to \mathbb P^1$ of degree $d$ is non-empty and smooth of dimension $2d+1-g$ as soon as $d\geq g+1$. But there are explicit cases for smaller $d$ where the corresponding space is singular. You can read about these results in this article by Akaohori and Namba.

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Here are some remarks about the last part of the question.

I wonder if the component of the space of maps $Hom(Y,Y)$ of degree $1$ and higher is not always smooth (it is not clear straight away how to consturct a contre-example). In the case of degree $1$ maps the space is clearly smooth since it is a Lie group of dimension $H^0(TY)$. In general, manifolds that admit slef-maps of degree higher than $1$ are not so common. For example if you take in $\mathbb CP^n$ ($n>3$) a hypersurface of degree $2$ and higher it does not admit self-maps of degree higher than $1$, this result is discussed in a nice article of Beauville

On the other hand, it is not hard to construct manfiolds for which some irreducible components of the space of self-maps of zero degree will be non-smooth. It is sufficient to take $Y$ such that $Hom(\mathbb CP^1,Y)$ is non smooth, and conisder $X=\mathbb CP^1\times Y$. Then you just conisder maps $X\to \mathbb CP^1\to X$.

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In fact, I meant \mathrm{Hom}(X,Y) in the last par of my question (I have edited it now), but looking at self-maps is fine. However, you claim that \mathrm{Hom}_1(Y,Y) is Lie group (for composition, I guess). Why is it so? (Sorry to be stupid, but I don't see why it's a group and why it's a manifold). – Oblomov May 31 '10 at 16:03
I have assumed, that Y is compact. In this case any self-map of degree one is an automorphism. Indeed, if we had a holomorphic map $Y\to Y$ of degree one that is not an automorphism, its differential would degenerate at one point and hence on a divisor. This would give a contradicton if we compaire the canoncial class of $Y$ with itself (sorry for been a bit clumsy). So these degree one maps form a group. – Dmitri May 31 '10 at 17:19
Dmitri, I suppose $Y$ is connected (via def'n of "variety"; always confuses me). Quasi-finite maps between smooth connected proj. varieties (over any field) of the same dimension are finite flat of constant degree, and finite flat maps with constant degree 1 between noetherian schemes are isom (pass to affine case and use Nakayama to reduce to considering fiber algebras of degree 1 over fields). To make a subfunctor of endomorphism functor out of the condition "finite flat of degree 1", we check it is an open condition on the Hom-scheme (e.g., by usual flatness, properness, fiber-dim. stuff). – BCnrd Jun 1 '10 at 2:31
BCnrd, thanks a lot for the suggestion of how to prove that the degree 1 End is an Auto! I wonder if the other thing that I said has any chances to be true -- for a smooth compact Y the spaces Hom(Y,Y) of degree>0 are smooth? – Dmitri Jun 1 '10 at 10:15

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