Tangent Spaces in Algebraic Geometry

We have discussed the notion of a tangent space in Differentiable Manifolds Revisited in the context of differential geometry. In this post we take on the same topic, but this time in the context of algebraic geometry, where it is also known as the Zariski tangent space (when no confusion arises, however, it is often simply referred to as the tangent space).

This will present us with challenges, since the concept of the tangent space is perhaps best tackled using the methods of calculus, but in algebraic geometry, we want to have a notion of tangent spaces in cases where we would not usually think of calculus as being applicable, for instance in the case of varieties over finite fields. In other words, we want our treatment to be algebraic. Nevertheless, we will use the methods of calculus as an inspiration.

We don’t want to be too dependent on the parts of calculus that make use of properties of the real and complex numbers that will not carry over to the more general cases. Fortunately, if we are dealing with polynomials, we can just “borrow” the “power rule” of calculus, since that “rule” only makes use of algebraic procedures, and we need not make use of sequences, limits, and so on. Namely, if we have a polynomial given by

$\displaystyle f=\sum_{j=1}^{n}ax^{j}$

We set

$\displaystyle \frac{\partial f}{\partial x}=\sum_{j=1}^{n}jax^{j-1}$

We recall the rules for partial derivatives – in the case that we are differentiating over some variable $x$ , we simply treat all the other variables as constants, and follow the usual rules of differential calculus. With these rules, we can now make the definition of the tangent space at the point $P$ with coordinates $(a_{1},a_{2},...,a_{n})$ as the algebraic set which satisfies the equation

$\displaystyle \sum_{j}\frac{\partial f}{\partial x_{j}}(P)(x_{j}-a_{j})=0$

For example, consider the parabola given by the equation $y-x^{2}=0$ . Let us take the tangent space at the point $P$ with coordinates $x=1$ , $y=1$ . The procedure above gives us

$\displaystyle \frac{\partial f}{\partial x}(P)(x-1)+\frac{\partial f}{\partial y}(P)(y-1)=0$

Since

$\displaystyle \frac{\partial f}{\partial x}=-2x$

$\displaystyle \frac{\partial f}{\partial y}=1$

We then have

$\displaystyle -2x|_{x=1,y=1}(x-1)+1|_{x=1,y=1}(y-1)=0$

$\displaystyle -2(1)(x-1)+1(y-1)=0$

$\displaystyle -2x+2+y-1=0$

$\displaystyle y-2x+1=0$

The parabola is graphed (its real part, at least, using the Desmos graphing calculator) in the diagram below in red, with its tangent space, a line, in blue:

desmos-graph

In case the reader is not convinced by our “borrowing” of concepts from calculus and claiming that they are “algebraic” in the specific case we are dealing with, another way to look at things without making reference to calculus is the following procedure, which comes from basic high school-level “analytic geometry”. First we translate the coordinate system so that the origin is at the point $P$ where we want to take the tangent space. Then we simply take the “linear part” of the polynomial equation, then translate again so that the origin is where it used to be originally. This gives the same results as the earlier procedure (the technical justification is given by the theory of Taylor series). More explicitly we have:

$\displaystyle y-x^{2}=0$

Translating the origin of coordinates to the point $x=1$ , $y=1$ , we have

$\displaystyle (y+1)-(x+1)^{2}=0$

$\displaystyle y+1-(x^{2}+2x+1)=0$

$\displaystyle y+1-x^{2}-2x-1=0$

$\displaystyle y-x^{2}-2x=0$

We take only the linear part, which is

$\displaystyle y-2x=0$

And then we translate the origin of coordinates back to the original one:

$\displaystyle (y-1)-2(x-1)=0$

$\displaystyle y-1-2x+2=0$

$\displaystyle y-2x+1=0$

which is the same result we had earlier.

But it may happen that the polynomial has no “linear part”. In this case the tangent space is the entirety of the ambient space. However, there is another related concept which may be useful in these cases, called the tangent cone. The tangent cone is the algebraic set which satisfies the equations we get by extracting the lowest degree part of the polynomial, which may or may not be the linear part. In the case that the lowest degree part is the linear part, the tangent space and the tangent cone coincide, and if this holds for all points of a variety, we say that the variety is nonsingular.

To give an explicit example, consider the curve $y^{2}=x^{3}+x^{2}$ , as seen in the diagram below in red (its real part graphed once again using the Desmos graphing calculator):

desmos-graph (2)

The equation that defines this curve has no linear part. Therefore the tangent space at the origin consists of all $x$ and $y$ which satisfy the trivial equation $0=0$ ; but then, all values of $x$ and $y$ satisfy this equation, and therefore the tangent space is the “affine plane” $\mathbb{A}^{2}$ . However, the lowest order part is $y^{2}=x^{2}$ , which is satisfied by all points which also satisfy either of the two equations $y=x$ or $y=-x$ . These points form the blue and orange diagonal lines in the diagram. Since the tangent space and the tangent cone do not agree, the curve is singular at the origin.

We can also define the tangent space in a more abstract manner, using the concepts we have discussed in Localization. Let $\mathfrak{m}$ be the unique maximal ideal of the local ring $O_{X,P}$ , and let $\mathfrak{m}^{2}$ be the product ideal whose elements are the sums of products of elements of $\mathfrak{m}$ . The quotient $\mathfrak{m}/\mathfrak{m}^{2}$ is then a vector space over the residue field $k$ . The tangent space of $X$ at $P$ is then defined as the dual of this vector space (the vector space of linear transformations from $\mathfrak{m}/\mathfrak{m}^{2}$ to $k$ ). The vector space $\mathfrak{m}/\mathfrak{m}^{2}$ itself is called the cotangent space of $X$ at $P$ . We can think of its elements as linear polynomial functions on the tangent space. There is an analogous abstract definition of the tangent cone, namely as the spectrum of the graded ring $\oplus_{i\geq 0}\mathfrak{m}^{i}/\mathfrak{m}^{i+1}$ .