First of all, I must thank you for taking the time to read this, since it is actually a question with multiple parts, and it's kind of long.

I am reading Kock and Vainsencher's An Invitation to Quantum Cohomology. In the very first page of Chapter 2 on Stable Maps, the following definition is given: (we work over $\mathbb{C}$)

Definition: By the degree of a map $\mu:\mathbb{P}^1\rightarrow \mathbb{P}^r$ we mean the degree of the direct image cycle $\mu_*[\mathbb{P}^1]$. In particular, a constant map has degree zero. In other words, if $e\geq 1$ is the degree of the image curve (with reduced scheme structure), and $k$ denotes the degree of the field extension corresponding to the map, then the degree of the map is $k\cdot e$. Note that, except for the case in which the image curve is a straight line, the definition above differs from the usual definition, given just by the degree of the field extension.

Then, the following is statement follows immediately:

To give a map $\mu:\mathbb{P}^1\rightarrow \mathbb{P}^r$ of degree $d$ is to specify, up to a constant factor, $r+1$ binary forms of degree $d$, which are not allowed to vanish simultaneously at any point. This condition defines a Zariski open subset $$ W(r,d) \subset \mathbb{P}\left(\bigoplus_{i-1}^r H^0(\mathcal{O}_{\mathbb{P}^1}(d))\right) $$

Now, for a book that claims to only require Chapter 1 of Hartshorne, all the above seem very cryptic. Specifically, my questions are the following:

  1. From what I have read on various other sources, the direct image cycle mentioned comes from the following map: $$ \mu_*: Z_k(\mathbb{P}^1) \rightarrow Z_k(\mathbb{P}^r) $$ of the abelian groups of $k$ cycles, however, I am struggling to find consistent definition on the constructions of these groups and the above map, like do we only consider 1-cycles in our case because $\mathbb{P}^1$ is irreducible? More specifically, how is the direct image cycle actually defined? And what is meant by its degree?
  2. How does the definition imply what follows "in other words..." about the degree of the field extension corresponding to the map?
  3. Which field extension is the field extension corresponding to the map? Is it $[\mathbb{C}(\mathbb{P}^r):\mathbb{C}(\mu(\mathbb{P}^1))]$? or $[\mathbb{C}(\mathbb{P}^1):\mathbb{C}(\mathbb{P}^r):]$? (I've seen both in different sources, specifically, the first one from Shafarevich and Danilov's Algebraic Geometry I: Algebraic Curves, Algebraic Manifolds and Schemes, Chapter II on Algebraic varieties and schemes Section 5; the second one from Hartshorne Chater II Section 6, pg. 137) Do either of them make sense?
  4. What exactly is the "usual definition"? And how does it differ from the definition given here?

  5. How does the definition given connect with the statement following it (the one on binary forms)? In particular what are all the objects in the inclusion statement? What is this $\mathbb{P}\left(\bigoplus_{i-1}^r H^0(\mathcal{O}_{\mathbb{P}^1}(d))\right)$ ?

Basically, this entire page does not make much sense to me, and I suppose all of the above points would be resolved from a complete explanation of the definition and what comes after it. My understanding of algebraic geometry is roughly at around beginning of Chapter 2 of Hartshorne on sheaves (As I mentioned before the author indicates that knowledge of Chapter 1 is sufficient, and one does not lose anything by considering schemes as varieties).

Any help is appreciated, and thanks for taking the time to read all this!

  • $\begingroup$ You have me confused since usually the extension covered by most books is wrt the function field of $[\mathbb{K}(V_1):\mathbb{K}(\mu^*(V_1))]$ for a map between varieties $V_1\to{}V_2$. $\endgroup$ Apr 1, 2019 at 16:09
  • $\begingroup$ @ΜάρκοςΚαραμέρης well that is actually part of my question which is to clarify what exactly IS the definition of the extension with respect to a map, and I suppose in particular which extension the author of this book is using $\endgroup$ Apr 1, 2019 at 16:11
  • $\begingroup$ In the general case when you have a morphis $\mu:A\to{}B$ It induces the pull back morphism $\mu^*=f(\mu)$ between their function fields $K(B)\to{}K(A)$. This is an injection since it is a homomorphism of fields and thus you can define the corresponding extension. $\endgroup$ Apr 1, 2019 at 16:36
  • $\begingroup$ I made a typo in the first comment that must have really confused you, apologies I meant $\mathbb{K}(V_1)\to\mu^*(\mathbb{K}(V_2))$ $\endgroup$ Apr 1, 2019 at 17:18

1 Answer 1

  • If you're more comfortable with singular homology, then I think it's helpful to think of singular homology instead. In this case the pushforward map is just the usual pushfoward map in homology $H_*(\mathbb{P}^1,\mathbb{Z}) \rightarrow H_*(\mathbb{P}^r,\mathbb{Z})$. If you want to stick with talking about cycles, then the (proper) pushforward map is just defined by mapping an irreducible subvariety $[V]$ to its image $[f(V)]$. Note that you need the properness assumption here to ensure that $f(V)$ is closed. It is a theorem that this induces a map on Chow groups $A_*$, so in this case we have $A_1(\mathbb{P}^1) \rightarrow A_1(\mathbb{P}^r)$.
  • The usual definition of the degree of a map between two varieties of the same dimension $V_1 \rightarrow V_2$ is given by the degree of the corresponding field extension.
  • There is another notion of a degree - this is defined for subvarieties of $\mathbb{P}^r$. Given a subvariety $V\subset \mathbb{P}^r$ of dimension $k$, its degree is defined to be the number of points you get after interesting $V$ with $n-k$ generic hyperplanes. For intuition, think in the case of curves in $\mathbb{P}^2$. Intersecting a degree $d$ curve (a curve cut out by a degree $d$ homogeneous equation) with a line gives $d$ points.
  • The definition of degree here is the product of the above two notions of degree. You can think of this number as the number of points in the pre-image of the set of number of points you get by intersecting your curve with a generic hyperplane.
  • So for example, the closed embedding $\mathbb{P}^1 \hookrightarrow \mathbb{P}^2$ as a conic is a map of degree $2$. The map $\mathbb{P}^1 \rightarrow \mathbb{P}^1$ given by $[X:Y] \mapsto [X^2:Y^2]$ is a map of degree $2$. If you compose these two maps, you get a map $\mathbb{P}^1 \rightarrow \mathbb{P}^1 \hookrightarrow \mathbb{P}^2$ of degree $4$. Your image is a conic. Intersecting the conic with a hyperplane gives 2 points. Each point has 2 points in the pre-image (well, generically) in the first map (the 2:1 cover).
  • This notion of degree, for a map $f:\mathbb{P}^1 \rightarrow \mathbb{P}^r$, also turns out to be also equal to $c$, where $f_*[\mathbb{P}^1] = c[L]$, where $[L] \in H_2(\mathbb{P}^r,\mathbb{Z})$ is the generator (the class of a line). You can replace $H_2$ with $A_1$ if you like. (Note that $H_2(\mathbb{P}^r,\mathbb{Z})$ / $A_1(\mathbb{P}^r)$ is isomorphic to $\mathbb{Z}$). In fact this is probably how I would define it in the first place.
  • A map from $\mathbb{P}^1 \rightarrow \mathbb{P}^r$ is always given by a $r+1$ tuple of binary forms of the same degree. To see that the degree of such a bilinear form has to be equal to $d$, you can use for example the equivalent definition above that the pre-image of the intersection of the image of your curve and a generic hyperplane should have $d$ points.

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