uni/ws2022/rav/lecture/rav11.tex

\documentclass{lecture}

\begin{document}

\section{The tangent cone and the Zariski tangent space}

\subsection{The tangent cone at a point}

Let $X \subseteq k^{n}$ be a non-empty Zariski-closed subset.

Let $P \in k[T_1, \ldots, T_n]$ be a polynomial. For all $x \in k^{n}$, we have a Taylor expansion
at $x$: For all $h \in k^{n}$:
\begin{salign*}
    P(x+h) &= P(x) + P'(x)h + \frac{1}{2} P''(x) (h, h) + \underbrace{\ldots}_{\text{finite number of terms}} \\
&= \sum_{d=0}^{\infty} \frac{1}{d!} P^{(d)}(x) (\underbrace{h, \ldots, h}_{d \text{ times}})
.\end{salign*}

\begin{bem}[]
    The term $\frac{1}{d!} P^{(d)}(x)$ is a homogeneous polynomial of degree $d$
    in the coordinates of $h = (h_1, \ldots, h_n)$:
    \begin{salign*}
    P^{(d)}(x) (h, \ldots, h)
    &= \sum_{\alpha \in \N_0^{n}, |\alpha| = d} \frac{d!}{\alpha_1! \cdots \alpha_n!}
    \frac{\partial^{|\alpha|}}{\partial T_1^{\alpha_1} \cdots \partial T_n^{\alpha_n}} P(x)
    h_1^{\alpha_1} \cdots h_n^{\alpha_n}
    .\end{salign*}

    Also, when $x = 0_{k^{n}}$ and if we write
    \[
    P = P(0) + \sum_{d=1}^{\infty} Q_d
    \] with $Q_d$ homogeneous of degree $d$, then for all $h = (h_1, \ldots, h_n) \in k^{n}$, we
    have
    \[
    \frac{1}{d!}P^{(d)}(0) \cdot (h, \ldots, h) = Q_d(h_1, \ldots, h_n)
    .\]
    For all $P \in \mathcal{I}(X) \setminus \{0\} $, we denote by
    $P_x^{*}$ the \emph{initial term} in the Taylor expansion of $P$ at $x$, i.e.
    the term $\frac{1}{d!} P^{(d)}(x) \cdot  (h, \ldots, h)$ for the smallest
    $d \ge 1$ such that this is not zero. If $P = 0$, we put $P_x^{*} \coloneqq 0$.
\end{bem}

\begin{definition}
    We set $\mathcal{I}(X)_x^{*}$ to be the ideal generated
    by $P_x^{*}$ for all $P \in \mathcal{I}(X)$.
\end{definition}

%\begin{satz}
%    The set $\mathcal{I}(X)_x^{*}$ is an ideal of $k[T_1, \ldots, T_n]$.
%\end{satz}
%
%\begin{proof}
%    By definition, $0 \in \mathcal{I}(X)_x^{*}$. Let $P_x^{*}, Q_x^{*}$ be elements
%    of $\mathcal{I}(X)_x^{*}$ coming from $P, Q \in \mathcal{I}(X)$. Then
%    $P_x^{*} - Q_x^{*}$ is of the form $R_x^{*}$ for some $R \in \mathcal{I}(X)$, where
%    $R = 0$, $R = P$, $R = Q$ or $R = P-Q$. Moreover, for $Q \in k[T_1, \ldots, T_n]$,
%    we have $P_x^{*} Q = (PQ)_x^{*} \in \mathcal{I}(X)_x^{*}$.
%\end{proof}

\begin{bem}[]
    The ideal $\mathcal{I}(X)^{*}$ is finitely generated. However,
    if $\mathcal{I}(X) = (P_1, \ldots, P_m)$, it is not true in general that
    $\mathcal{I}(X)_x^{*} = ((P_1)_x^{*}, \ldots, (P_m)_{x}^{*})$. We may need
    to add the initial terms at $x$ of some other polynomials of the
    form $\sum_{k=1}^{m} P_k Q_k \in \mathcal{I}(X)$.

    If $\mathcal{I}(X) = (P)$ is principal though, we have $\mathcal{I}(X)_x^{*}
    = (P_x^{*})$.
\end{bem}

\begin{definition}
    The \emph{tangent cone} to $X$ at $x$ is the affine algebraic
    set
    \[
    \mathcal{C}_x^{(X)} \coloneqq x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})
    = \{ x + h \colon h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})\} 
    .\] 
\end{definition}

\begin{bem}
    The algebraic set $\mathcal{C}_x(X)$ is a cone at $x$: It contains $x$ and
    for all $x + h \in \mathcal{C}_x(X)$ for some $h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})$,
    we have for all
    $\lambda \in k^{\times}$,
    $\lambda h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*}$), i.e. $x + \lambda h \in
    \mathcal{C}_x(X)$.

    Indeed, $P_x^{*} \in \mathcal{I}(X)_x^{*}$ is either zero or a homogeneous polynomial of
    degree $r \ge 1$. Thus for $h \in k^{n}$ and $\lambda \in k^{\times}$:
    $P_x^{*}(\lambda h) = \lambda^{r} P_x^{*}(h)$ which
    is $0$ if and only if $P_x^{*}(h) = 0$.
\end{bem}

\begin{bsp}[]
    Let $k$ be an infinite field and let $P \in k[x,y]$ be an irreducible polynomial
    such that $X \coloneqq \mathcal{V}(P)$ is infinite. Then we know that
    $\mathcal{I}(X) = (P)$. Then we can determine $\mathcal{C}_X(X)$ by computing
    the successive derviatives of $P$ at $x$: In this case
    $\mathcal{I}(X)_x^{*} = (P_x^{*})$. For convenience wie will mostly consider examples
    for which $x = 0_{k^2}$.
    \begin{enumerate}[(i)]
        \item $P(x,y) = y^2 - x^{3}$. Then $P^{*}_{(0,0)} = y^2$, so the tangent cone
            at $(0, 0)$ is the algebraic set
            \[
            \mathcal{C}_{(0,0)}(X) = \{ (x,y) \in k^2  \mid y^2 = 0\} 
            .\]

        \begin{figure}[h]
            \centering
            \begin{tikzpicture}
                \begin{axis}[
                    legend style={at={(0.02, 0.98)}, anchor=north west}
                    ]
                \algebraiccurve[red]{y^2 - x^3}
                \algebraiccurve[green][$y^2 = 0$]{y}
                \algebraiccurve[blue][$y = \frac{3}{2}x - \frac{1}{2}$]{y-1.5*x + 0.5}
                \end{axis}
            \end{tikzpicture}
            \caption{The green line is the tangent cone at $(0,0)$ and the blue line
            the tangent cone at $(1,1)$.}
        \end{figure}
        Note that $P_{(1,1)}^{*}(h_1, h_2) = 2h_2 - 3 h_1$, so the tangent cone at
        $(1,1)$ is
        \begin{salign*}
            \mathcal{C}_{(1,1)}(X) &= \{ (1 + h_1, 1 + h_2)  \mid 2h_2 - 3h_1 = 0\} \\
                                   &= \left\{ (x,y) \in k^2  \mid y = \frac{3}{2} x - \frac{1}{2}\right\} 
        .\end{salign*}
    \item $P(x,y) = y^2 - x^2(x+1)$. Then $P_{(0,0)}^{*} = y^2 - x^2$ so
        \[
        \mathcal{C}_{(0,0)}(X) = \{ y^2 - x^2 = 0\} 
        \] which
        is a union of two lines.
        \begin{figure}[h]
            \centering
            \begin{tikzpicture}
                \begin{axis}[
                    legend style={at={(0.02, 0.98)}, anchor=north west}
                    ]
                \algebraiccurve[red][$y^2 = x^2(x+1) $]{y^2 - x^2*(x+1)}
                \algebraiccurve[green]{y^2 - x^2}
                \end{axis}
            \end{tikzpicture}
            \caption{The green line is the tangent cone at $(0,0)$.}
        \end{figure}

        In contrast, $P_{(1,1)}^{*}(h_1, h_2) = 2h_2 - 5h_1$ so
        \[
        \mathcal{C}_{(1,1)}(X) = \left\{ (x,y) \in k^2  \mid y = \frac{5}{2} x - \frac{3}{2}\right\}
        ,\] which is just one line.
        Evidently this is related to the origin being a ,,node`` of the curve of equation
        $y^2 - x^2(x+1) = 0$.
    \end{enumerate}
\end{bsp}

\begin{bem}
    \begin{enumerate}[(i)]
        \item The tangent cone $\mathcal{C}_x(X)$ represents all directions coming out
            of $x$ along which the initial term $P_x^{*}$
            vanishes, for all $P \in \mathcal{I}(X)$. In that sense, it is the least complicated
            approximation to $X$ around $x$, in terms of the degrees of the polynomials involved.
        \item The notion of tangent cone at a point enables us to define singular points of algebraic
            sets and even distinguish between the type of singularities:
            Let $\mathcal{I}(X) = (P)$.

            When $\text{deg}(P_x^{*}) = 1$, the tangent cone to $X \subseteq k^{n}$ at $x$
            is just an affine hyperplane, namely $x + \text{ker } P'(x)$, since
            $P_x^{*} = P'(x)$ in this case. The point $x$ is then called \emph{non-singular}.

            When $\text{deg}(P_x^{*}) = 2$, we say that $X$ has a \emph{quadratic singularity}
            at $x$. If $X \subseteq k^2$, a quadratic singularity is called a \emph{double point}.
            In that case,
            $P_x^{*} = \frac{1}{2} P''(x)$ is a quadratic form on $k^2$. If it is non-degenerate,
            then $x$ is called an \emph{ordinary} double point. For instance,
            if $X$ is the nodal cubic of equation $y^2 = x^2(x+1)$, then the origin is
            an ordinary double point (also called a \emph{node}), since
            $\frac{1}{2}P''(0,0)$ is the quadratic form associated to the symmetric matrix
            $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} $.
            But if $X$ is the cuspidal cubic of equation $y^2 = x^{3}$, then
            the origin is \emph{not} an ordinary double point, since
            $\frac{1}{2}P''(0,0)$ corresponds to $\begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} $.
            Instead, the origin is a \emph{cusp} in the following sense. We can write
            \[
            P(x,y) = l(x,y)^2 + Q_3(x,y) + \ldots
            \] with $l(x,y) = \alpha x + \beta y$ a linear form in $(x,y)$, and the double point
            $(0,0)$ is called a cusp if $Q_3(\beta, -\alpha) \neq 0$. This means that
            \[
            t ^{4}X P(\beta t, - \alpha t)
            \] in $k[t]$. And this is indeed what happens for $P(x,y) = y^2 - x^{3}$, since
            $l(x,y) = y$ and $Q_3(x,y) = -x^{3}$.
    \end{enumerate}
\end{bem}

\begin{bem}[]
    One can define the \emph{multiplicity} of a point $(x,y) \in \mathcal{V}_{k^2}(P)$ as
    the smallest integer $r \ge 1$ such that $P^{(r)}(x,y)\neq 0$.
    If $P^{(r)}(x,y) \cdot (h, \ldots, h) = 0 \implies h = 0_{k^2}$, the singularity
    $(x,y)$ is called \emph{ordinary}. If $k$ is algebraically closed and
    $(x,y) = (0,0)$, we can write
    $P^{(r)}(0, 0) = \prod_{i=1}^{m} (\alpha_i x + \beta_i y)^{r_i} $,
    with $r_1 + \ldots + r_m = r$. Then $(0,0)$ is an ordinary singularity of multiplicity $r$
    iff $r_i = 1$ for all $i$. For instance, $(0,0)$ is an ordinary triple point of the trefoil
    curve $P(x,y) = (x^2 + y^2)^2 + 3x^2 y - y^{3}$.
\end{bem}

\subsection{The Zariski tangent space at a point}

Let $X \subseteq k^{n}$ be a Zariski-closed subset and $x \in X$.

The tangent cone is in general not a linear approximation. To remedy this, one can
consider the Zariski tangent space to $X$ at a point $x \in X$.

\begin{definition}
    The \emph{Zariski tangent space} to $X$ at $x$ is the affine subspace
    \[
    T_xX \coloneqq x + \bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x)
    .\]
\end{definition}

\begin{bem}[]
    By translation, $T_xX$ can be canonically identified to the vector space
    $\bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x) $.
\end{bem}

\begin{satz}[]
    View the linear forms
    \[
    P'(x) \colon h \mapsto P'(x) \cdot h
    \] as homogeneous polynomials of degree $1$ in the coordinates of $h \in k^{n}$ and
    denote by
    \[
    \mathcal{I}(X)_x \coloneqq (P'(x) : P \in \mathcal{I}(X))
    \] the ideal generated by these polynomials. Then
    \[
    T_xX = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x)
    .\]
\end{satz}

\begin{proof}
    It suffices to check that
    \[
    \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) = \bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x)
    \]
    which is obvious because the $(P'(x))_{P \in \mathcal{I}(X)}$ generate $\mathcal{I}(X)_x$.
\end{proof}

\begin{korollar}
    $T_xX \supseteq \mathcal{C}_x(X)$
    \label{kor:cone-in-tangent-space}
\end{korollar}

\begin{proof}
    Since $\mathcal{I}(X)_x \subseteq \mathcal{I}(X)_x^{*}$, one has
    $\mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) \supseteq \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})$.
\end{proof}

\begin{definition}
    If $T_xX = \mathcal{C}_x(X)$, the point $x$ is called \emph{non-singular}.
\end{definition}

\begin{satz}
    If $\mathcal{I}(X) = (P_1, \ldots, P_m)$, then
    $\mathcal{I}(X)_x = (P_1'(x), \ldots, P_m'(x))$
\end{satz}

\begin{proof}
    By definition,
    \[
        (P_1'(x), \ldots, P_m'(x)) \subseteq (P'(x) : P \in \mathcal{I}(X)) = \mathcal{I}(X)_x
    .\] But for $P \in \mathcal{I}(X)$, there exist $Q_1, \ldots, Q_m \in k[T_1, \ldots, T_n]$ such
    that $P = \sum_{i=1}^{m} Q_i P_i$, so
    \begin{salign*}
        P'(x) &= \sum_{i=1}^{m} (Q_i P_i)'(x) \\
              &= \sum_{i=1}^{m} (Q_i'(x) \underbrace{P_i(x)}_{= 0} + \overbrace{Q_i(x)}^{\in k}
              P_i'(x))
    \end{salign*}
    since $x \in X$. This proves that $P'(x)$ is in fact a linear combination of the linear
    forms $(P_i'(x))_{1 \le i \le m}$.
\end{proof}

\begin{korollar}
    If $\mathcal{I}(X) = (P_1, \ldots, P_m)$, then
    $T_xX = x + \bigcap_{i=1}^{m} \operatorname{ker } P_i'(x)$.
    Moreover, if we write $P = (P_1, \ldots, P_m)$, and view this
    $P$ as a polynomial map $k^{n} \to k^{m}$, then
    \[
    T_xX = x + \text{ker } P'(x)
    \] with $P'(x)$ the Jacobian of $P$ at $x$, i.e.
    \[
        P'(x) = \begin{pmatrix} \frac{\partial P_1}{\partial T_1}(x) & \cdots & \frac{\partial P_1}{\partial T_n}(x) \\
        \vdots & & \vdots \\
        \frac{\partial P_m}{\partial T_1}(x) & \cdots & \frac{\partial P_m}{\partial T_n}(x)
    \end{pmatrix} 
    .\] In particular, $\text{dim } T_xX = n - \operatorname{rk } P'(x)$.
    \label{kor:tangent-kernel-jacobian}
\end{korollar}

\begin{bsp}
    \begin{enumerate}[(i)]
        \item $X = \{ y^2 - x^{3} = 0\} \subseteq k^2$. Then $\mathcal{I}(X) = (y^2 - x^{3})$,
            so,
            \[
            T_{(0,0)}X = (0,0) + \text{ker} \begin{pmatrix} 0 & 0 \end{pmatrix} = k^2
            .\] 
            which strictly contains the tangent cone $\{y^2 = 0\} $. In particular,
            the origin is indeed a singular point of the cuspidal cubic. In general,
            \[
            T_{(x,y)}X = (x,y) + \text{ker} \begin{pmatrix} -3x^2 & 2y \end{pmatrix} 
            ,\] 
            which is an affine line if $(x,y) \neq (0,0)$.
        \item $X = \{ y^2 - x^2 - x^{3} = 0\} \subseteq k^2$. Then
            $\mathcal{I}(X) = (y^2 - x^2 - x^{3})$, so
            \[
                T_{(0,0)}X = (0,0) + \text{ker} \begin{pmatrix} 0 & 0 \end{pmatrix} = k^2
            \] which again strictly contains the tangent cone $\{y = \pm x\} $. In general,
            \[
                T_{(x,y)}X = (x,y) + \text{ker} \begin{pmatrix} -2x & 2y \end{pmatrix}
            ,\] which is an affine line if $(x,y) \neq (0,0)$.
    \end{enumerate}
\end{bsp}

\begin{bem}
    The dimension of the Zariski tangent space at $x$ (as an affine subspace of $k^{n}$)
    may vary with $x$.
\end{bem}

%\begin{satz}[a Jacobian criterion]
%    If $(P_1, \ldots, P_m)$ are polynomials such that
%    $\mathcal{I}(X) = (P_1, \ldots, P_m)$ and $\operatorname{rk } P'(x) = m$, where
%    $P = (P_1, \ldots, P_m)$, then $x$ is a non-singular point of $X$.
%\end{satz}
%
%\begin{proof}
%    By \ref{kor:cone-in-tangent-space} and \ref{kor:tangent-kernel-jacobian} it suffices to show that
%    \[
%    \mathcal{C}_x(X) \supseteq x + \bigcap_{i=1} ^{m} \text{ker } P_i'(x)
%    .\] By definition
%    \[
%    \mathcal{C}_x(X) = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})
%    \] and $\mathcal{I}(X)_x^{*} = \{ Q_x^{*} : Q \in \mathcal{I}(X)\} $. If $Q \in \mathcal{I}(X)$,
%    there exist polynomials $Q_1, \ldots, Q_m$ such that
%    $Q = \sum_{i=1}^{m} Q_i P_i$, so $Q_x^{*}$ is a linear combination of the $(P_i)_x^{*}$.
%    Since $\text{rk }(P_1'(x), \ldots, P_m'(x)) = m$, we have
%    $P_i'(x) \neq 0$ for all $i$. So $(P_i)_x^{*} = P_i'(x)$ in the Taylor expansion
%    of $P_i$ at $x$. So $Q_x^{*}$ is a linear combination
%    of $(P_1'(x), \ldots, P_m'(x))$,
%    which proves that if $h \in \bigcap_{i=1}^{m} \text{ker } P_i'(x)$, then
%    $Q_x^{*}(h) = 0$ for all $Q \in \mathcal{I}(X)$, hence
%    $x + h \in \mathcal{C}_x(X)$.
%\end{proof}

\end{document}