Separating Points and Tangent Vectors

Suppose that $X/k$ is a scheme and that $\phi:X\to \mathbb{P}^n_k$ is a morphism. It’s not hard to see that $\phi^* \mathscr{O}(1)$ is an invertible sheaf and is generated by the sections $s_i=\phi^*x_i$ . Conversely, given an invertible sheaf $\mathscr{L}$ and global section $s_0,\ldots,s_n$ which globally generated $\mathscr{L}$ there is a morphism $\phi: X \to \mathbb{P}^n$ such that $\phi^* \mathscr{O}(1)=\mathscr{L}$ and $s_i=\phi^* x_i.$ This may seem like a complicated statement but the idea is really quite simple.

To see the simplicity, let’s look at the simple case of $\mathscr{O}(2)$ on $\mathbb{P}^1$ , and take the global sections $x^2,xy,y^2.$ The map we are talking about above is simply the map from $\mathbb{P}^1 \to \mathbb{P}^2$ given by $(x,y)\mapsto (x^2,xy,y^2).$ (I don’t know how to do square brackets on wordpress.) In fact the only thing we really have to check to make sure this is well defined, is that it doesn’t have any base points, i.e. points which get sent to $(0:0:0).$ It’s easy to see this is not the case here, but what more important is that the intuition here is exactly the same in the general case. The system $\{\mathscr{L},s_i\}$ will have base points (in this sense) if and only if the sections do not generate the sheaf. So for example, the sections $\{x^2,xy\}$ do not globally generate, because they do not generate the sheaf at the point $(0,1)$ . If you really want to work out the details, this all boils down to the fact that locally at this point, we can divide by $y$ since it is nonzero, but we cannot divide by $x$ so we will never get the function $y$ .

To construct the map $\phi$ in general you need to define maps locally and patch them together, but the basic idea is exactly the same as in this example. One useful application of this notion is the fact that automorphisms of $\mathbb{P}^n$ are precisely given by $(n+1)\times (n+1)$ matrices modulo scalar multiples. That such matrices yield automorphisms is just a simple exercise in matrix multiplication, which is good to do! To see the other direction, given a morphism $\phi:\mathbb{P}^n \to \mathbb{P}^n$ since Pic $\mathbb{P}^n\cong \mathbb{Z}$ and since $\phi$ is an automorphism, and $\mathscr{O}(1)$ is a generator, we know that $\phi^*\mathscr{O}(1)$ must be also be a generator. Thus it is $\mathscr{O}(\pm 1)$ and since only one of these has global sections, we conclude it must be $\mathscr{O}(1).$ Thus the pullbacks of the $x_i$ in the image must be linear combinations of the $x_i$ in the domain, and thus we can use these equations to form a matrix. (Forgive me for not being explicit here!) as required. Finally uniqueness follows since giving a morphism is the same as giving a sheaf (in this case $\mathscr{O}(1)$ and some sections (the linear combinations).

All of this generality shows us the existence of morphisms if we take a sheaf and some sections, but the map we end up with could be very badly behaved. For example, it could fail to be injective, and other horrors. The best case scenario would be if this map were a closed immersion (turning our scheme $X$ into a projective variety!) Luckily there’s a criterion for when this happens, and we’ll state it now.

Proposition: Let $k$ be an algebraically closed field, let $X$ be a projective scheme over $k$ and let $\phi: X\to \mathbb{P}^n$ be a morphism (over k) corresponding to $\{\mathscr{L},s_i\}$ as above. Let $V\subset \Gamma(X,\mathscr{L})$ be the subspace spanned by the $s_i$ . Then $\phi$ is a closed immersion if and only if

elements of V separate points, i.e. for any two distinct points $P,Q\in X$ there is an $s\in V$ such that $s\in m_p\mathscr{L}_p$ but $s\notin m_Q\mathscr{L}_Q$ , or vice versa, and
elements of V separate tangent vectors, i.e. for each closed point $P\in X$ , the set $\{s\in V~|s_P\in m_P\mathscr{L}_P \}$ spans the k-vector space $m_P\mathscr{L}_P/m_P^2\mathscr{L}_P.$

[I’m not going to give the proof here, but the idea is that the first condition implies that the morphism is injective (and we get that the map is a homeomorphism onto its image via standard nonsense about images of proper schemes etc) whereas the second condition is used to prove that the structure map is surjective. The algebraically closed condition is necessary to get a handle on the types of closed points.]

Instead, let’s see this theorem in action. Let’s use $\mathbb{P}^1$ and $\mathscr{O}(2)$ again. The vector space spanned by the sections $\{x^2,y^2\}$ has no base points, but it does not separate points, since $(-1,1),(1,1)$ both get sent to $(1,1)$ .

The notion of separating tangent vectors is a bit more subtle, but the geometric idea is that you don’t want to “squash” tangent vectors. For example, the 2-1 squashing map for the curve $x^2=yz$ where the map $C\to \mathbb{P}^1$ is given by $(x,y,z)\mapsto (y,z)$ squashes the origin. Indeed, if P is the origin, (0,0,1) then $m_P/m_P^2$ is generated by $x$ but neither of $y,z$ is a generator. In fact $y\in m_P^2$ and $z$ is a unit.

Finally, I’ll close with the somewhat simper statement for a curve $X.$ In this case, the tangent space is 1-dimensional, so for the set $\{s~|~s_P\in m_P \}$ to generate $m_P/m_P^2$ just says that every function which vanishes at $P$ does not vanish twice there. If you know about divisors this is just saying that $P$ is not a base point of $D-P$ if $D$ is the divisor associated to the map.

~ by twistedcubic on January 19, 2009.

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One Response to “Separating Points and Tangent Vectors”

Hey, how do you put those little LaTeX bits in your blog? You probably won’t see this, will you…

Hungrygrad said this on October 15, 2009 at 3:50 pm | Reply

The Twisted Cubic