# A multivariable substitution generalization.

I solved an integral using the variable substitution

$$x = \frac{u-v}{\sqrt{2}}$$ $$y = \frac{u+v}{\sqrt{2}}$$

This is interesting because the Jacobian of this transformation is $$1$$. Particularly, this transformation corresponds to a clockwise rotation of $$\pi/4$$ around the origin.

The problem involved an integrand containing $$xy$$, which is transformed to $$1/2(u^2-v^2)$$.

Now, out of curiosity, are there other transformations in higher dimensions sending $$x_1x_2 \cdots x_n$$ to $$a\cdot(u_1^2 \pm u_2^2 \pm \ldots \pm u_n^2)$$ for some constant $$a$$? What is the Jacobian, if such transformations exist? Do these transformations involve complex coefficients, and what are the transformations with real coefficients?

Edit: As J.G. pointed out in the comments, it should probably be $$a\cdot(u_1^n \pm u_2^n \pm \ldots \pm u_n^n)$$.

• Are you sure you didn't want the $u_j$ squared?
– J.G.
Commented Dec 25, 2023 at 14:07
• @J.G. yes, thank you. Commented Dec 25, 2023 at 14:11
• On second thoughts, they should probably be to the $n$th power.
– J.G.
Commented Dec 25, 2023 at 14:32
• @J.G. mhhh, probably yes. Also, it makes sense under a dimensional point of view. Commented Dec 25, 2023 at 14:36
• @Francesco The theory of quadratic forms allows you to transform any expression of the form $\sum_{i,j = 1}^n a_{ij}x_ix_j$ into the form $\sum_{i=1}^n a_i u_i^2$ with a suitable transformation, and with the spectral theorem we can always take this substitution to be an orthogonal transformation (which would have Jacobian $1$). I don't believe that this generalizes nicely to higher exponents. Commented Dec 25, 2023 at 22:17

There is no such substitution. Moreover, there is no differentiable substitution which transforms the product $$x_1 x_2 \cdots x_n$$ with $$n > 2$$ into a sum of the form $$f_1(u_1)+ \dots + f_n(u_n)$$ where the functions $$f_i$$ are such that the derivatives $$f_i'$$ have only finite number of zeros.
Let $$x=(x_1, \dots, x_n)$$, $$u = (u_1, \dots, u_n)$$, let $$P(x) = \prod_{i = 1}^n x_i$$ be the product, and let $$F(u)=\sum_{i = 1}^n f_i(u)$$.
Assume $$F(u) = P(X(u))$$ for some function $$X \colon \mathbb{R}^n \to \mathbb{R}^n$$. Note that if $$X_1(u)=0$$ and $$X_2(u)=0$$, then $$\nabla F(u) = \sum_{i = 1}^n \nabla X_i(u)\prod_{j \neq i} X_j(u)=0.$$ Thus we have $$\nabla F(u)=0$$ for all points $$u$$ corresponding to the points of the linear subspace $$x_1 = x_2 = 0$$. If $$n > 2$$, the number of such points is infinite. On the other hand, $$\nabla F(u) = 0$$ is just $$f_i'(u_i)=0$$ for all $$i$$, and because of our condition on $$f_i$$, the number of such $$u$$ is finite. We have a contradiction.
The condition on $$f_i$$ can be relaxed if instead of cardinality we compare dimension (codimension 2 for $$X_1(u) = X_2(u) = 0$$ and dimension 0 for $$f_1'(u_1) = \dots = f_n'(u_n)=0$$). My guess would be that the conditions on $$f_i$$ and $$X$$ can be relaxed to just continuity, but I don't know topology well enough to prove this.