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In Higher plane curves, nr 167, Salmon proves that the cross-ration of the four tangents to a non-singular plane cubic, drawn from a point on the curve, is independent of the point. A proof can be found in Van der Waerden's Einführung in die algebraische Geometrie. But can anyone explain in modern terms Salmon's proof? It runs as follows:

If from two consecutive points $O$, $O'$ of the curve we draw the two sets of tangents $OA$, $OB$, $OC$, $OD$; $O'A$, $O'B$, $O'C$, $O'D$, any tangent $OA$ intersects the consecutive tangent $O'A$ in its point of contact. Now the four points of contact $A$, $B$, $C$, $D$ lie on the polar conic of $0$, which also touches the cubic at the point (Art. 64) ; hence the six points $OO'ABCD$ lie on the same conic, and therefore the anharmonic ratio of the pencil $\{O.ABCD\}$ is the same as that of the pencil $\{O'.ABCD\}$. Since then this ratio remains the same when we pass from one point of the curve to the consecutive one, we learn that the anharmonic ratio is constant of the pencil formed by the four tangents which can be drawn from any point of the curve.

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Could you give the precise reference? In the version of the text available at michigan universiy, I did not find it at page 167 and neither at article 167 (which is at page 156 an talk about polars and tangents, but not about this question) –  Jérémy Blanc Feb 22 at 13:31
    
@Jérémy , I referred to the 3rd edition, from 1879, see archive.org/details/117724690. The Michigan text is the first edition, where it is Art 158 on page 150. –  Jan Stevens Feb 22 at 16:14
    
Thanks. The site you indicate is nice. It seems to me that the points of contact $A,B,C,D$ are really the points such that the tangent at $A$ passes through $O$. Hence, they cannot be the same for two distinct points. Is this a mistake or we dont get something? –  Jérémy Blanc Feb 22 at 17:24
    
Of course the points are not distinct. The best I can think of is that you should view a curve as the envelope of its tangents. Wikipedia says: Classically, a point on the envelope can be thought of as the intersection of two "adjacent" curves, meaning the limit of intersections of nearby curves. –  Jan Stevens Feb 22 at 18:29
    
But if you take two points which are the same (or one infinitely near), I dont see how you could prove that the cross-ratio does not depend on the point. –  Jérémy Blanc Feb 22 at 20:10
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2 Answers

up vote 3 down vote accepted

Consider a local parametrization $O(t), A(t), ...$, with $O = O(0), A = A(0), ...$. Salmon is essentially checking that $\frac{d}{dt}(O(t)A(t),O(t)B(t);O(t)C(t),O(t)D(t)) = 0$ at $t = 0$, where $(OA,OB;OC,OD)$ denotes the anharmonic ratio of the pencil $\{O.ABCD\}$.

Since $A(t)$ stays on the line $OA$ to first order, and similarly for $B(t), C(t), D(t)$, we have

$\frac{d}{dt}(O(t)A(t),O(t)B(t);O(t)C(t),O(t)D(t))|_{t=0} = \frac{d}{dt}(O(t)A,O(t)B;O(t)C,O(t)D)|_{t=0}$.

This will be zero if and only if $O(t)$ stays on the conic $OABCD$ to first order, which Salmon checks using the fact that for $O$ on the cubic curve the polar conic of $O$ passes through $O, A, B, C, D$ and is tangent to the cubic curve at $O$.

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The way I see this result is the following.

Fix a line $L \cong \mathbb{P}^1$ and a point $x \in X$, where $X \subset \mathbb{P}^2$ is the smooth plane cubic.

The projection $\pi_x \colon X \longrightarrow L$ is a double cover, branched in four points corresponding to the four tangent lines drawn from $x$ to $X$. The cross-ratio of these four points is equal to the cross ratio $\lambda(x)$ of the four tangents. On the other hand, the isomorphism class of the double cover is completely determined by the $j$-invariant $$j:= 2^8\frac{(\lambda^2- \lambda + 1)^3}{\lambda^2(\lambda-1)^2}.$$ Since $X$ is fixed, this means that the image of the holomorphic map $$\lambda \colon X \longrightarrow \mathbb{C}, \quad x \mapsto \lambda(x)$$ is a finite set of values. But then $\lambda$ is constant.

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This is too modern. As far as I understand, the proof in question is a step in establishing the well-definedness of the $j$-invariant of a cubic: the $j$ invariant of the ramification locus is independent of the center of projection. What I don't understand is what "consecutive points" are and how we get same four tangents from two distinct points. –  Alex Degtyarev Feb 21 at 15:03
    
Consecutive means that O' is infinitely near to O. –  Jan Stevens Feb 21 at 15:27
    
I really cannot follow Salmon's proof. The only points where the first polar of $X$ with respect to $O$ intersects $X$ are the four points $A$, $B$, $C$, $D$ (with multiplicity $1$) and the point $O$ (with multiplicity $2$, in fact the polar conic is tangent to $X$ at $O$). In other words, it seems to me that necessarily $O=O'$. I cannot give a meaning to "infinitely near" in this context, either. Maybe I'm missing something... –  Francesco Polizzi Feb 21 at 16:08
    
I agree with you that I do not understand how it is possible to have four tangents points A,B,C,D associated to two points $O$, $O'$. In fact, from a point on the curve with obtain four points of tangent and conversely each point of tangent gives back $O$. There is thus some kind of directed graph, with one arrow starting from each vertex in one direction and 4 in the other. Also, infinitely near seems strange here. –  Jérémy Blanc Feb 21 at 16:56
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Actually to talk about the cross-ratio you need to order the 4 lines -- I doubt that this can be done globally. The reasonable approach is to take the "invariant cross-ratio", namely the $j$-invariant as Francesco does. –  abx Feb 21 at 17:20
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