# Integration question

Say you have a function f(w, x, y, z) = (w[sup]2[/sup] - x[sup]2[/sup] - y[sup]2[/sup] - z[sup]2[/sup], 2wx, 2wy, 2wz). Now we all know what the derivative of this function is, but is there any sense in which it’s integrable? My best guess is that it looks a lot like a gradient, and so we could find a function F such that grad(F) = f. What’s the official approach to a problem like this?

Wow, it’s hard to give a succinct answer to this one. This is one of the major topics in multivariate calculus (e.g., Stokes’ theorem and its special cases), so that might make a good starting point. On the other hand, many multivariate calculus books are especially interested in the case of 3 dimensions, which tends to obscure some important issues. It really helps to know some tensor analysis (which is largely just multivariate calculus without the assumptions of 3-dimensional Cartesian space).

That said, it sounds like you want to find a scalar field F whose gradient is a given vector field f. There’s a well-known three-dimensional result: such F exists iff f is curl-free, dxf = 0 (using d for the nabla operator). This does generalize to higher dimensions and more general manifolds, but doing so requires more care than in three dimensions. We can no longer blithely switch between “covariant” and “contravariant” vectors, and the curl of a 1-form (a covariant vector field) is not a vector field but a 2-form. (The equation for the curl works out just as you’d expect, though.)

What this works out to mean is fairly simple. f is a gradient 1-form iff df[sub]i[/sub]/dx[sup]j[/sup]=df[sub]j[/sub]/dx[sup]i[/sup] (using “d” as the partial derivative operator). This is not the case for the example you gave: e.g., df[sub]w[/sub]/dx=-2x, df[sub]x[/sub]/dw=2x (off by a sign).

So why did I bother with all of that tensor calculus stuff? Well, in the presence of a metric it’s fairly common to switch somewhat blithely between covariant and contravariant vectors, raising or lowering indexes via the metric tensor g. Your 4-dimensional example looks like it might be in the Minkowski metric (I’ll say signature (-1,1,1,1) here); if so, if f is actually a contravariant vector, then because of the sign switch it actually is curl-free, and so it is a gradient (namely, of w(x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup])-w[sup]3[/sup]/3).

Does this make any sense? I’m sure I’ve used improper terminology somewhere here… With a little more detail on the properties of f and F it might be possible to give more specific answers.

This is a pretty standard sort of problem in multivariate calc (what I wish I was teaching now).

We want:

dF/dw = w[sup]2[/sup] - x[sup]2[/sup] - y[sup]2[/sup] - z[sup]2[/sup]
dF/dx = 2wx
dF/dy = 2wy
dF/dz = 2wz

so, check the second partials.

d[sup]2[/sup]F/dwdx = -2x = -d[sup]2[/sup]F/dxdw
d[sup]2[/sup]F/dwdy = -2y = -d[sup]2[/sup]F/dydw
d[sup]2[/sup]F/dwdz = -2x = -d[sup]2[/sup]F/dzdw
d[sup]2[/sup]F/dxdy = 0 = d[sup]2[/sup]F/dydx
d[sup]2[/sup]F/dxdz = 0 = d[sup]2[/sup]F/dzdx
d[sup]2[/sup]F/dydz = 0 = d[sup]2[/sup]F/dzdy

So, there cannot be a potential function F.

In terms of differential forms, we have a 1-form p with

p = (w[sup]2[/sup] - x[sup]2[/sup] - y[sup]2[/sup] - z[sup]2[/sup]) dw + 2wx dx, 2wy dy+ 2wz dz

d(p) = -2x dx^dw - 2y dy^dw - 2z dz^dw + 2x dw^dx + 2y dw^dy + 2z dw^dz
= 4x dw^dx + 4y dw^dy + 4z dw^dz != 0

so there cannot be an F with dF=p, since d(dF) = dp should be 0.

Unfortunately, I’m not as familiar with tensors as I should be, so I can’t weigh in on that aspect of the discussion. The function I chose is perhaps a bit far out–in quaternionic terms, f(z) = z[sup]2[/sup], which sure seems like it ought to be integrable over some sets (no idea which). What about something nicer, like the analogous function over C? Then we have f(x, y) = (x[sup]2[/sup] - y[sup]2[/sup], 2xy). Is there a notion for the integral of that without appealing to complex analysis?

If this comes from quaternions, I actually know why this doesn’t work, since I’m currently taking Igor Frenkel’s quaternionic analysis course.

Basically, if you try to use the limit-of-difference-quotient definition of derivative, you only get linear functions to be differentiable. If you use the analogue of \bar{d}f = 0 you get more, but the set of regular functions doesn’t form a ring, and the identity function isn’t even regular. Basically, in quaternions things turn very bad very quickly.

As for complex analysis, note that the operator you need to use is 1/2 (d/dx - i d/dy). If you approach it as a multivariate problem, (x[sup]2[/sup] - y[sup]2[/sup]) dx + (2xy) dy doesn’t have a vanishing exterior derivative, so there’s no potential function.

The problem is that there’s two different notions of “derivative” going on here. Using {1,i} as a basis and d/dz as the operator makes the function “integrable”. Using {dx,dy} as a basis and considering the exterior derivative d doesn’t.