4
$\begingroup$

I asked this on Math.StackExchange, but received no response, so trying here ...

A paper I'm reading says the following ...

With homogeneous coordinates $\mathbf{x} = [x,y,z,w]$, let $F(\mathbf{x}) = 0$ be the equation of a surface of degree $n$. The first polar form of $F(\mathbf{x})$ at the pole $\mathbf{a} = [a,b,c,h]$ is defined as $$ F^1(\mathbf{x}) = \frac1n(aF_x + bF_y + cF_z + hF_w) $$

And then, later, there's a similar definition ...

Let $\mathbf{p}(x)$ be a polynomial curve of degree $n$ and let $\mathbf{p}(x,w)$ denote its homogeneous form. Its first polar with respect to the pole $x_1$ is defined by $$ \mathbf{p}^1(x_1 \,|\,x) = \frac1n(x_1\mathbf{p}_x + \mathbf{p}_w) $$

The paper says these concepts are "well-known in algebraic geometry". I'm having trouble understanding what these things mean geometrically. Could someone explain, or provide a reference, please. I'd like the explanation or reference to be something simple and concrete in 2D or 3D space, please; abstraction and generality probably won't help me.

I recall polar lines of circles and conics from high school analytic geometry. That's the case $n=2$, and it makes sense. It's the $n > 2$ case that I'm having trouble grasping.

I'm guessing that there is some relationship with "polar forms" (aka blossoms in the CAGD field). In this context, the polar form of a polynomial $f:\mathbb{R} \to \mathbb{R}$ of degree $n$ is a symmetric multi-affine function $F:\mathbb{R}^n \to \mathbb{R}$ such that $F(x, \ldots,x) = f(x)$. But I can't see that relationship, either.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ The relation is $F^1(x) = F(a,x,\dots,x)$. For a discussion, see "Polar covariants of plane cubics and quartics" by Dolgachev and Kanev. $\endgroup$
    – Sasha
    Commented Apr 2, 2016 at 5:48

1 Answer 1

Reset to default
8
$\begingroup$

Geometrically, one first meets polar hypersurfaces when studying tangent lines from a point to a hypersurface. In fact, this concept generalizes the classical polarity of a point with respect to a conic.

More precisely, let $X \subset \mathbb{P}^k$ be a hypersurface of degree $n$ given by the zero locus of a homogeneous polynomial $F$, namely $$F(x_0, \ldots, x_k)=0,$$ and let $\mathbf{a}=[a_0: \ldots:a_k] \in \mathbb{P}^k$ be any point. We define the first polar of the point $\mathbf{a}$ with respect to $X$ as the hypersurface $P_{\mathbf{a}}(X)$ of degree $n-1$ defined by $$\sum_{i=0}^k a_i\frac{\partial F}{\partial x_i}=0$$ (dividing by $n$ is not really important here).

Then the locus $X \cap P_\mathbf{a}(X)$ is given by the points $\mathbf{p}$ such that the tangent space to $X$ at $\mathbf{p}$ contains $\mathbf{a}$.

In the particular case where $X \subset \mathbb{P}^2$ is a smooth plane curve of degree $n$, the first polar is a curve of degree $n-1$. Therefore we deduce from the Bézout theorem that for a general point $\mathbf{a} \in \mathbb{P}^2$ pass exactly $n(n-1)$ distinct tangent lines to $X$. This number is classically called the class of $X$. When $X$ is singular the formula for the class must be modified according to the number and the local type of the singular points; in the case of nodes and cusps, this leads to the so-called Plücker formulas.

Dolgachev's excellent book Classical Algebraic Geometry probably contains all you need about this subject and much more.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Thanks. The goal of Dolgachev's book is to rephrase the classical results in a modern way, which makes things more difficult for me. But he provides lots of pointers into the classical literature, which will be useful, I expect. $\endgroup$
    – bubba
    Commented Apr 2, 2016 at 9:21
  • 1
    $\begingroup$ The second displayed equation in your answer says (to me) that the vector $\mathbf{a} = (a_1, \ldots, a_n)$ is orthogonal to the surface normal $\mathbf{n}(x) = \big(\partial F/ \partial x_1, \ldots , \partial F/ \partial x_n\big)$. In other words, it says $\langle \mathbf{a}, \mathbf{n}(x) \rangle = 0$. From my experience with conics and quadrics, I was expecting something like $\langle \mathbf{x} - \mathbf{a}, \mathbf{n}(x) \rangle = 0$, instead, saying that the line from $\mathbf{a}$ to $\mathbf{x}$ is tangent to the surface at the point $\mathbf{x}$. $\endgroup$
    – bubba
    Commented Apr 2, 2016 at 9:30
  • 2
    $\begingroup$ We are working in the projective setting, not in the affine setting. If $\mathbf{b}:=[b_0: \ldots : b_n]$ is a smooth point of $X$, that is $F(b_0, \ldots, b_n)=0$, then the equation of the tangent hyperplane of $X$ at $\mathbf{b}$ is $$\sum_{i=0}^n \frac{\partial F}{\partial x_i}(\mathbf{b})x_i=0.$$ Therefore the second displayed equations tells you precisely that if $\mathbf{b}$ belongs to the intersection of $X$ and the first polar $P_{\mathbf{a}}(X) $, then the tangent hyperplane of $X$ at $\mathbf{b}$ contains $\mathbf{a}$. $\endgroup$
    – Francesco Polizzi
    Commented Apr 2, 2016 at 13:58
  • 1
    $\begingroup$ I see. Not used to thinking projectively. Thanks. $\endgroup$
    – bubba
    Commented Apr 9, 2016 at 13:38
  • 1
    $\begingroup$ Snce $F$ is homogeneous of degree $d$ it satisfies $\sum_{i=0}^n \frac{\partial F}{\partial x_i} x_i = dF$. In particular if $F(\mathbf{x})=0$ then $\sum (\mathbf{x}_i-\mathbf{a}_i) \frac{\partial F}{\partial x_i}(\mathbf{x}_i) = - \sum \mathbf{a}_i \frac{\partial F}{\partial x_i}(\mathbf{x}_i)$, and the minus sign doesn't change what hypersurface is defined by this equation. So I think homogeneity of $F$ is what matters here. $\endgroup$
    – Zach Teitler
    Commented May 10, 2016 at 8:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .