Tangent Spaces and Smoothness

As in differential geometry, in algebraic geometry the tangent space is a key tool for understanding the local structure of a variety.

Definition of the Tangent Space

In differential geometry, for the sheaf \(\mathcal{C}^\infty_M\) of smooth functions defined on \(M\), we confirmed that the collection of all germs vanishing at a point \(x\in M\),

\[\mathfrak{m}_x=\{\mathbf{f}\in \mathcal{C}^\infty_x\mid \mathbf{f}(x)=0\}\]

is a maximal ideal. We then proved that the tangent space can be viewed as

\[(\mathfrak{m}_x/\mathfrak{m}_x^2)^\ast\]

([Manifolds] §Cotangent Space, ⁋Lemma 1). This process is usually not covered much in differential geometry, but it is of great help in generalizing to algebraic varieties. Namely, (fixing the affine case for convenience) we already know what functions defined on algebraic varieties are (§Quasi-Projective Varieties, ⁋Definition 5), and we also know that in this case, the collection of all functions vanishing at \(x\in X\) corresponds to the maximal ideal of \(\mathbb{K}[X]\) corresponding to this point. Therefore, we define this as

\[\mathfrak{m}_x=\{f\in \mathbb{K}[X]\mid f(x)=0\}\]

and can consider the localization \(\mathbb{K}[X]_{\mathfrak{m}_x}=\mathcal{O}_{X,x}\) of \(\mathbb{K}[X]\) at this maximal ideal. ([Commutative Algebra] §Localization, ⁋Definition 4) Geometrically, keeping §Affine Varieties, ⁋Definition 14 in mind, these can be defined as the germs of regular functions at the point \(x\).

Definition 1 We define the Zariski tangent space \(T_x X\) of a variety \(X\) at a point \(x\) by

\[T_x X = (\mathfrak{m}_x / \mathfrak{m}_x^2)^\ast\]

where \(\mathfrak{m}_x\) is the unique maximal ideal of the local ring \(\mathcal{O}_{X,x}\) at the point \(x\).

Ultimately, the key point of this definition is that since the quotient \(\mathfrak{m}_x / \mathfrak{m}_x^2\) contains the first-order infinitesimal data at \(x\), we define it as the Zariski cotangent space \(T_x^\ast X\). And its dual \(T_x X\) is the space of linear functionals acting on this data, i.e., directional derivative operators, and this definition agrees with defining \(T_xX=\Der_\mathbb{K}(\mathcal{O}_{X,x}, \mathbb{K})\).

We do not use analysis-style differentiation of the \(\epsilon\)-\(\delta\) form, but varieties are essentially defined by polynomials, and their differentiation can be understood formally: differentiating \(\x^n\) yields \(n\cdot \x^{n-1}\). In particular, in the case of affine varieties, this can be written down more explicitly.

Proposition 2 Consider an affine variety \(X = Z(f_1, \ldots, f_k) \subseteq \mathbb{A}^n\) and its corresponding ideal \(I(X) = (f_1, \ldots, f_k)\). Then at a point \(x = (x_1, \ldots, x_n)\) of \(X\),

\[T_x X \cong \{v \in \mathbb{K}^n \mid (\dd{f_i})_x(v) = 0 \text{ for all } i\}\]

Here, \((\dd{f_i})_x\) is the differential of \(f_i\) at \(x\), given by

\[(\dd{f_i})_x(v) = \sum_{j=1}^n \frac{\partial f_i}{\partial \x_j}(x) v_j\]

Proof

Consider the coordinate ring \(\mathbb{K}[X] = \mathbb{K}[\x_1, \ldots, \x_n] / (f_1, \ldots, f_k)\) of \(X\). Since \(\mathfrak{m}_x = (\x_1 - x_1, \x_2 - x_2, \ldots, \x_n - x_n) / (f_1, \ldots, f_k)\), we have

\[\mathfrak{m}_x / \mathfrak{m}_x^2 \cong (\x_1 - x_1, \x_2 - x_2, \ldots, \x_n - x_n) / \left( (\x_1 - x_1, \x_2 - x_2, \ldots, \x_n - x_n)^2 + (f_1, \ldots, f_k) \right)\]

Taylor expanding each \(f_i\) at \(x\) gives

\[f_i = \sum_{j=1}^n \frac{\partial f_i}{\partial \x_j}(x) (\x_j - x_j) + \text{higher order terms}\]

and the higher order terms lie in \((\x_1 - x_1, \x_2 - x_2, \ldots, \x_n - x_n)^2\). Thus, in \(\mathfrak{m}_x / \mathfrak{m}_x^2\), the linear parts \(\sum_j \frac{\partial f_i}{\partial \x_j}(x) (\x_j - x_j)\) of the \(f_i\) become zero.

On the other hand, \(\mathfrak{m}_x / \mathfrak{m}_x^2\) is generated by linear combinations of the \(\x_j - x_j\), so it can be viewed as a quotient of \(\mathbb{K}^n\). The kernel of the differential \((\dd{f_i})_x\) then corresponds exactly to the directions that vanish in \(\mathfrak{m}_x / \mathfrak{m}_x^2\). Taking the dual, we obtain

\[T_x X = (\mathfrak{m}_x / \mathfrak{m}_x^2)^\ast \cong \{v \in \mathbb{K}^n \mid (\dd{f_i})_x(v) = 0 \text{ for all } i\}\]

Although the proof is written in a complicated way using the language of maximal ideals, the underlying philosophy is simple if one considers \(X=Z(f_i)\). In this case, \((\dd{f_i})_x(v)=0\) (if we view \(\mathbb{K}^n\) as \(\mathbb{A}^n\)) is precisely the (ordinary) tangent space of the hypersurface \(Z(f_i)\) in \(\mathbb{A}^n\). Although Proposition 2 by itself applies only to affine varieties, since any point \(x\) of an arbitrary variety \(X\) has an affine neighborhood, it essentially applies to all varieties. The same is true of the following proposition on the dimension of the tangent space.

Proposition 3 \(T_x X\) is a \(\mathbb{K}\)-vector space, and its dimension is \(n - \rank(J_x)\). Here, \(J_x\) is the \(k \times n\) Jacobian matrix

\[J_x = \left(\frac{\partial f_i}{\partial \x_j}(x)\right)_{1 \le i \le k, 1 \le j \le n}\]

Proof

Each \((\dd{f_i})_x: \mathbb{K}^n \rightarrow \mathbb{K}\) is a linear functional. By Proposition 2, \(T_x X\) is the intersection of their kernels, hence a subspace of \(\mathbb{K}^n\). The rows of the Jacobian matrix \(J_x\) are the coordinate expressions of these linear functionals, so

\[T_x X = \ker(J_x) = \{v \in \mathbb{K}^n \mid J_x v = 0\}\]

By the rank-nullity theorem, \(\dim T_x X = n - \rank(J_x)\).

Smooth Points and Singular Points

In differential geometry, the dimension of the tangent space at any point was always equal to the dimension of the manifold. However, this is because the definition of a manifold is rather restrictive; in algebraic geometry, even an affine variety defined by a single polynomial may fail to be a manifold (in the classical picture). (Example 6 (Smooth points), Example 7 (Singular points)) Nevertheless, the dimension of the tangent space and the dimension of the variety are not completely unrelated.

Proposition 4 For an irreducible variety \(X\) and any point \(x\), we have \(\dim T_x X \ge \dim X\).

Proof

We prove only the affine case. Let \(X = Z(f_1, \ldots, f_k) \subseteq \mathbb{A}^n\) be irreducible with \(\dim X = d\). At the point \(x \in X\), consider the local ring \(\mathcal{O}_{X,x} = \mathbb{K}[X]_{\mathfrak{m}_x}\). Since \(X\) is irreducible, \(\mathbb{K}[X]\) is a finitely generated \(\mathbb{K}\)-algebra and a domain, and since \(\mathfrak{m}_x\) is a maximal ideal, \(\dim \mathbb{K}[X]/\mathfrak{m}_x = 0\). Therefore, the dimension formula of [Commutative Algebra] §Noether Normalization, ⁋Theorem 4 gives \(\codim \mathfrak{m}_x = \dim \mathbb{K}[X]\), and since the codimension of a prime ideal is defined as the dimension of its localization ([Commutative Algebra] §Krull Dimension, ⁋Definition 2), we obtain the following equation:

\[\dim \mathcal{O}_{X,x} = \codim \mathfrak{m}_x = \dim \mathbb{K}[X] = \dim X = d\]

Here the third equality follows from §Dimension, ⁋Proposition 2.

In general, for a Noetherian local ring \((R, \mathfrak{m})\), we have \(\dim_{\mathbb{K}}(\mathfrak{m}/\mathfrak{m}^2) \ge \dim R\). ([Commutative Algebra] §System of Parameters, ⁋Proposition 2) Therefore, we have

\[\dim T_x X = \dim_{\mathbb{K}}(\mathfrak{m}_x/\mathfrak{m}_x^2) \ge \dim \mathcal{O}_{X,x} = d = \dim X\]

To build our intuition, it is useful to examine when the inequality holds. We call such points singular points.

Definition 5 A point \(x \in X\) is called a smooth point (or nonsingular point) if \(\dim T_x X = \dim X\). Otherwise (i.e., if \(\dim T_x X > \dim X\)), it is called a singular point.

Example 6 (Smooth points)

1. Every point of \(\mathbb{A}^n\) is a smooth point. Since \(\mathbb{A}^n\) has no defining equation, \(T_x \mathbb{A}^n = \mathbb{K}^n\), and \(\dim T_x \mathbb{A}^n = n = \dim \mathbb{A}^n\).
2. Every point of the parabola \(Z(\y - \x^2)\) is a smooth point. For \(f = \y - \x^2\), we have \(J_{(x,y)} = (-2x, 1)\), which is nonzero at every point. Therefore, \(\dim T_x X = 2 - 1 = 1 = \dim X\).

Example 7 (Singular points)

1. (Node) Consider \(X = Z(\y^2 - \x^2(\x+1)) \subseteq \mathbb{A}^2\). This curve splits into two branches at the origin.

   The Jacobian of this curve is

   \[J_{(x,y)} = \begin{pmatrix} -2x - 3x^2 & 2y \end{pmatrix}\]

   so at the origin the Jacobian is \((0,0)\), and thus by Proposition 3 the origin is a singular point. Geometrically, the fact that the tangent space is 2-dimensional means that both tangent directions of the two branches are included. Concretely, since \(\y^2 - \x^2(\x+1) \approx \y^2 - \x^2 = (\y-\x)(\y+\x)\), near the origin the curve looks like the union of the two lines \(\y = \x\) and \(\y = -\x\). The node is one of the “mildest” singular points.

2. (Cusp) Now consider \(Z(\y^2 - \x^3)\subseteq \mathbb{A}^2\).

   Here, the origin of this curve is a singular point. To verify this, computing the Jacobian gives

   \[J_{(x,y)}=\begin{pmatrix}-3x^2&2y\end{pmatrix}\]

   so at the origin we have \(\nabla f(0,0) = (0, 0)\). Geometrically, the fact that the tangent space is 2-dimensional in this situation means that every direction is “tangent” at the origin, which indicates that the curve is too sharp to define a tangent line in any direction.

In the examples above, we naturally used the following proposition.

Proposition 8 (Jacobian criterion) Consider an affine variety \(X = Z(f_1, \ldots, f_k) \subseteq \mathbb{A}^n\) and its corresponding ideal \(I(X) = (f_1, \ldots, f_k)\). Then a point \(x \in X\) is a smooth point if and only if the rank of the Jacobian matrix \(J_x\) is \(n - \dim X\).

Proof

In Proposition 3, we showed that \(\dim T_x X = n - \rank(J_x)\). In Definition 5, that \(x\) is a smooth point means that \(\dim T_x X = \dim X\). Therefore, \(x\) is a smooth point if and only if

\[n - \rank(J_x) = \dim X\]

that is, \(\rank(J_x) = n - \dim X\).

Existence of Smooth Points

Any algebraic variety is smooth at most of its points. To show this, we need the notion of a generic point.

Definition 9 For an irreducible variety \(X\), the generic point \(\eta\) is the unique point belonging to every nonempty open subset of \(X\).

In the affine case \(X = \Spec A\), \(\eta\) corresponds to the minimal prime ideal of \(A\) (i.e., the \((0)\) ideal), and the local ring \(\mathcal{O}_{X,\eta}\) is precisely the function field \(K(X) = \Frac(A)\). Geometrically, the generic point can be thought of as the “most general point” of \(X\), a point having no particular property of \(X\). We can use this idea in the following proof.

Proposition 10 For an irreducible variety \(X\), the set \(X_\sm\) of smooth points is a dense open subset of \(X\). In particular, \(X_\sm \ne \emptyset\).

Proof

Let the dimension of \(X = Z(f_1, \ldots, f_k) \subseteq \mathbb{A}^n\) be \(\dim X = d\). By Proposition 8 (Jacobian criterion),

\[X_\sm = \{x \in X \mid \rank(J_x) = n - d\}\]

holds. We now show that this set is a dense open subset. First, that \(X_\sm\) is an open set is relatively clear. By Proposition 3 and Proposition 4, at any point of \(X\), \(n - \rank(J_x) = \dim T_x X \ge d\), that is, \(\rank(J_x) \le n-d\) holds automatically, so on \(X\) the condition that the rank is exactly \(n-d\) is equivalent to the condition that the rank is at least \(n-d\). But the latter is equivalent to the determinant of some \((n-d) \times (n-d)\) submatrix not being \(0\), which is an open condition in the Zariski topology. Therefore, \(X_\sm\) is an open subset of \(X\).

Showing that \(X_\sm\) is nonempty is somewhat technical; the idea is that since the rank of the Jacobian matrix \(J=(...)\) over the function field \(K(X)\) is \(n-d\), a suitable \((n-d)\times (n-d)\) minor is nonzero in \(K(X)\), and the (nonempty) open set where this minor does not vanish is contained in \(X_\sm\). Therefore, \(X_\sm\) is nonempty, and any nonempty open subset is dense by irreducibility.

We then define the following.

Definition 11 A variety \(X\) is called smooth (or nonsingular) if every point is a smooth point, i.e., \(X_\sm = X\). Otherwise (i.e., if a singular point exists), it is called singular.

Example 12 The varieties in Example 6 (Smooth points) are all smooth, and all the varieties in Example 7 (Singular points) are singular.

Tangent Cone

At a singular point, the tangent space is too large to accurately reflect the local structure of the variety. In this case, the tangent cone provides more precise information. Intuitively, the tangent space being too large means that the rank of the Jacobian is too small, which happens, for example, because the linear approximation of the given function alone provides no information. Therefore, if we consider higher-degree approximations of the given function, the situation might change.

To this end, for any polynomial \(f\in \mathbb{K}[\x_1,\ldots, \x_n]\), we define the initial term of \(f\), \(\initial(f)\), to be the homogeneous component of \(f\) of smallest degree. Then, for any ideal \(\mathfrak{a}\), we define the initial ideal of \(\mathfrak{a}\), \(\initial(\mathfrak{a})\), to be the homogeneous ideal generated by the \(\initial(f)\).

Definition 13 For any affine variety \(X\subseteq \mathbb{A}^n\), we define the algebraic variety defined by \(\initial(I(X))\) to be the tangent cone of \(X\) at the origin, and denote it by \(TC_0 X\).

More generally, by writing \(f\) as a polynomial in the \(\x_i-x_i\) and making an analogous definition, one can define the tangent cone at an arbitrary point. The reason this is called a cone is that, just as in §Projective Varieties, ⁋Definition 12, it is the zero set of a homogeneous ideal.

Let us now see how this gives a finer classification of singular points.

Example 14 For the nodal curve \(X = Z(\y^2 - \x^2(\x+1))\) in Example 7 (Singular points), the lowest degree term of \(f\) is \(\y^2 - \x^2 = (\y-\x)(\y+\x)\), so

\[TC_0 X = Z(\y-\x) \cup Z(\y+\x)\]

This precisely shows that the node splits in the directions of the two lines \(\y = \x\) and \(\y = -\x\).

Example 15 For the curve \(X = Z(\y^2 - \x^3)\) in Example 7 (Singular points), the lowest degree term of \(f\) is \(\y^2\), so

\[TC_0 X = Z(\y^2)\]

This is the line \(\y = 0\) counted twice, and it shows that the cusp ends sharply in the \(\x\)-axis direction. By comparison, the tangent space \(T_0 X = \mathbb{K}^2\) is too large, containing all directions.

In general, keeping §Rational Maps, ⁋Example 12 in mind, the singularity of a nodal curve can be resolved by blowup. That is, after blowing up, the two branch lines \(\y-\x\) and \(\y+\x\) at the origin are separated by \(\mathbb{P}^1\). However, for a cusp the curve does not split into two branches near the origin, so even after blowup the point lying over the origin is only a single point of \(\mathbb{P}^1\); in this sense one generally regards a cusp as a worse singularity than a node.

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References

[Har] J. Harris, Algebraic Geometry: A First Course, Springer, 1992.
[Sha] I. R. Shafarevich, Basic Algebraic Geometry I: Varieties in Projective Space, Springer, 2013.