## Double erratum on double operations

I made an extremely silly error in class last week that I attempted to correct hastily, but then ended up making a mistake even in the correction. I am deeply indebted to Chris Smith who pointed out the initial error, and to Alex Tao for noticing the `corrected’ error.

The context of the problem was the quadratic form defined in class as

q(v)=(d^2/dt^2)f(p+tv)|t=0

Here, f is a smooth function defined on the plane and the vector space is that of arrows based as a point p. So we are restricting the function to the line that goes through p in the direction v, and then taking the second derivative at p (corresponding to t=0).

The error in question arose when we were discussing the case where f is the distance-squared from a fixed line L. We wished to find the real canonical form of q. First, let me present the correct solution. We let v be a vector orthogonal to the line and w a vector parallel to the line. We will compute the matrix of q with respect to the basis v and w. Let d be the distance between p and L. Then clearly,

f(p+tv)=(d+t||v||)^2=d^2+2dt||v||+t^2||v||^2

so that

q(v)=2||v||^2.

Meanwhile, since f is constant along the line p+tw, we have q(w)=0. Let’s compute q(v+w). The point is that

f(p+t(v+w))=f(p+tv+tw)

is the same as f(p+tv). It would be nice at this point to draw a picture, but my graphic skills are very limited. Imagine walking from p to p+tv, and then from there to p+tv+tw. Clearly, along the second portion, you will be walking parallel to L, and hence, the distance will remain unchanged. As a result,

q(v+w)=[(d^2/dt^2)f(p+t(v+w))|t=0]=[(d^2/dt^2)f(p+tv)|t=0]=q(v),

and therefore

q(v+w)-q(v)-q(w)=q(v)-q(v)=0.

Thus, the matrix we want is

{(2||v||^2 , 0), (0,0)}

which has real canonical form

{(1,0),(0,0)}

since 2||v||^2 is positive.

In my (poor) defense, my mistakes could be interpreted as the weakness of turning problems into pure algebra. Of course, algebra is a powerful tool, as I emphasized in my `remark on quadratic forms.’ But then, since the symbolic manipulations have no intuitive meaning, it’s easy to make mistakes. On the other hand, this is what makes algebraic manipulations very important when seeking the aid of a computer, for example.

Now to the key point of my error: I made it seem like I could derive a real canonical form

{(1,0),(0,0)}

just from the fact that q(v)>0 and q(w)=0. That this is nonsense can be seen by considering the form

{(1,0),(0,-1)}.

In this case,

q(1,0)>0

while

q(1,1)=0.

Also, as Alex correctly pointed out, any form

{(0,b),(b,0)}

with b non-zero has real canonical form

{(1,0),(0,-1)}

I’ll leave this as an exercise rather than risk another mistake 🙂

MK