# Cotangent space of a variety

I encountered this definition of the contangent space of a variety $V$ at $x$,

$$T^*_x = \mathfrak m_x/\mathfrak m_x^2$$

with $\mathfrak m_x = \{f\in k[V]: f(x)=0 \}$. Could you help me with it?

From a differential point of view, we can define the (co-)/tangent space in a point by taking equivalence classes of curves going through that point. The definition above seems kind of similar, but I am not used to the algebraic setting.

First of all, I don't really understand the definition of $k[V]$. I know, that we can get it by restricting polynomials to $V$, but why is that $k[V]= k[x_1,\dots,x_n]/I$?

I want to get a better understanding of this language by going through some examples.

First I look at a sphere $S^2$ given by $x^2+y^2+z^2-1=0$. Then $p_x=2x$, $p_y=2y$ and $p_z=2z$. Thus $p_x=p_y=p_z=0$ only at $(0,0,0)$ but this is not on $S^2$. So for every $(a,b,c)\in S^2$ the tangent space is given by $$2a(x-a)+2b(y-b)+2c(z-c)=0,$$ which is $2$-dimensional.

Why does its dual correspond to $\mathfrak m_x/\mathfrak m_x^2$.

Another example is $V$ given by $$x^3+y^3-3xy=0,$$ then $p_x=3(x^2-y)$ and $p_y=3(y^2-x)$.
So $p_x=p_y=0$ at $(0,0)\in V$.

Hence $V$ has a singular point. How does this reflect in the abstract formalism?

• The way I always think of this is that $\mathfrak m_x$ is the maximal ideal of the local ring $\mathcal O_x$, consisting of all functions vanishing at $x$. In two variables, that would be all functions generated by $x$ and $y$. The second power $\mathfrak m_x^2$ is all functions generated by $x^2,xy,y^2$. When you take functions generated by $x$ and $y$ and reduce that modulo functions generated by $x^2, xy$ and $y^2$, you get combinations of just $x$ and $y$. That gives all linear functions. – Fly by Night Nov 30 '17 at 18:26

## 1 Answer

I think @fly by night and I have the same basic intuition for why $\mathfrak{m}/\mathfrak{m}^2$ is the cotangent space (see my answer here for basically a recapitulation of the comment.) But let's see how to see that its duel is the tangent space without this knowledge.

For question $1$: For me, the clarifying interpretation comes from the definition of the tangent space in terms of "derivations." For each $v \in T_p(X)$, we are allowed to take a directional derivative with respect to this vector. In this way, the vectors in the tangent space can be identified with linear functions $D:k[x,y,z]/(f) \to k$ ( recall that the quotient should be thought of as functions well defined on the sphere) that satisfy the leibniz rule $D(fg)=f(p)D(f)+g(p)D(p)$.

Given your definition of $\mathfrak{m}_p$, we can take the actual definition of ideal multiplication to be $$\mathfrak{m}_p^2:=\{\sum_{k} f_kg_k: f_k, g_k \in \mathfrak{m}_p \}.$$

From this, it is clear that by the Leibniz rule, every function $D$ will vanish on $\mathfrak{m}_p^2$, (these are functions that vanish at $p$ and whose derivatives fanish as well) making it the case that every $D \in T_p(X)$ will descend to a map $$D: \mathfrak{m}/\mathfrak{m}^2 \to k$$

which is exactly the dual of the cotangent space. The quotient has a nice heuristic in that it captures first order behaviors of functions on a variety, and it takes equivalence classes of functions up to their first order behavior by quotienting by $\mathfrak{m}^2$

For question 2: I have a blog post on this for curves, that might help.

However, you can form the local ring

$(K[x,y]/(x^3+y^3-3xy))_{(x,y)}$

and we can form the cotangent space, but $\tilde{x},\tilde{y} \in \mathfrak{m}/\mathfrak{m}^2$, and these two area linearly independent over $k$, so the dimension of the cotangent space is too large (it is $2$.)

I suppose this formula is reflecting the fact that for a polynomial,

$p(x)=F_m(x)+ F_{m+1}(x)+ \dots+ F_n(x)$, where $F_i$ are all homogeneous of degree $i$, we can see that we can define the multiplicity of $p$ at $(0,0)$ to be precisely $m$ (the least degree homogoenouos component.) Using the usual rules of deriviative, we need that $m>0$ for the curve to go through $(0,0)$ and the degree $m=1$ for the polynomial to be nonsingular. If $m>1$, $(0,0)$ is a double point, triple point, etc.

In your case, the $2$-dimensionality of the (co)tangent space is reflected in the fact that $$p(x)=(x^3+y^3)-(3xy)$$

so $m(p)=2$, so it is a double point and in fact the quadratic component factors into two lines $x=0$, $3y=0$ (which if you graph the function, you will see is correct.)

Moreover, this agrees with a more analytic flavor of argument since if we study $p(x)$ as $x,y \to (0,0)$ the least degree part of the polynomial dominates the rest, since they shrink rapidly, so $p(x,y) \sim -3xy$ near zero.

See chapter 3.2 of the curve book for the melding of the local ring and this notion of multiplicity.

• However, I invite others to give a better answer for number $2$, my computational abilities in this kind of thing are very rusty. – Andres Mejia Dec 9 '17 at 23:45