It can be checked that the Vandermonde determinant defined as $$V(\alpha_1, \cdots, \alpha_n) = \prod_{1 \le i < j \le n}(\alpha_i-\alpha_j) $$ is a harmonic function, that is $\Delta V = 0$ where $\Delta$ is the Laplace operator. Is there a deeper or more intuitive reason why this fact should hold? The straightforward proof of just computing the derivatives and checking doesn't provide any insights.

## 3 Answers

Consider the symmetric group action permuting the variables. The Vandermonde determinant $V$ is antisymmetric, meaning it spans an alternating representation—it's invariant under permutations, up to multiplication by the sign of the permutation. Applying any symmetric differential operator (such as the Laplacian) preserves antisymmetry, but lowers the degree (as long as the operator doesn't have any constant terms).

And $V$ is the lowest-degree antisymmetric form. This is a fun, quick exercise. First note that $\deg V = \binom{n}{2}$, which is $0 + 1 + \dotsb + (n-1)$, and indeed all the monomials appearing in $V$ have the form $x_1^0 x_2^1 \dotsm x_n^{n-1}$, up to permutation and coefficient of $\pm 1$. None of the exponents here are repeated, and we realize that in any lower degree polynomial, there isn't enough room to have distinct exponents. Now if $f$ is any antisymmetric polynomial with a term $c x_1^{a_1} \dotsm x_n^{a_n}$ with a repeated exponent $a_i = a_j$, then permuting by the transposition $(i \, j)$ leaves this term unchanged; but it has to take this term to $-c x_1^{a_1} \dotsm x_n^{a_n}$ in order for $f$ to be antisymmetric; so $c=0$. Only terms with pairwise-distinct exponents can appear in $f$, so $\deg f$ must be at least $\binom{n}{2}$.

This actually proves a bit more: up to scalar factor, $V$ is the unique antisymmetric polynomial of degree $\binom{n}{2}$, and in fact any antisymmetric polynomial is divisible by $V$. This also generalizes to other finite reflection groups. You can see, for example, Chapter 20 of Kane's book Reflection groups and invariant theory.

But at the moment we just care about the property of having minimal degree. Now the point is that applying a symmetric differential operator preserves the antisymmetry property, but lowers the degree. But the only lower-degree antisymmetric form is just zero.

I would argue that this approach provides insights: it generalizes to other reflection groups, which are deeply studied. For me, it came up in relation to apolarity and Waring rank, where it was useful to know which differential operators annihilate $V$. (The above shows that symmetric differential operators lie in the ideal of annihilators, and it turns out they generate the ideal.)

The harmonicity of $V$ can be understood and placed in a more general context by identifying $V$ as the Doob h-transform$^\ast$ of $n$ independent and identically distributed diffusion processes, see Orthogonal polynomial ensembles in probability theory page 433. The generalization is that ${\cal D}V=0$ when $${\cal D}=\sum_{i=1}^n\biggl[(ax_i+b)\frac{\partial^2}{\partial \alpha_i^2}+c\frac{\partial}{\partial\alpha_i}\biggr]$$ for some $a,b,c\in\mathbb{R}$. This covers the cases of Brownian motion, squared Bessel processes (squared norms of Brownian motions) and generalized Ornstein-Uhlenbeck processes driven by Brownian motion.

_{$^\ast$ A diffusion process of $\alpha_1,\alpha_2,\ldots\alpha_n$, starting at zero, conditioned on $\alpha_1<\alpha_2<\cdots<\alpha_n$.}

The other answers already answered the question. I just wanted to point out that this type of question is also part of the Dyson Brownian motion topic. You can see that from the answers already stated, for example Carlo Beenakker. If you look up Dyson Brownian motion you can get to this old blog post of Tao

https://terrytao.wordpress.com/2010/01/18/254a-notes-3b-brownian-motion-and-dyson-brownian-motion/

Around equation (9) he starts to address the type of algebra you mention. In your notation, his equation (11) says $$ \frac{\partial}{\partial \alpha_i} V\, =\, \sum_{\substack{j\\j\neq i}} \frac{V}{\alpha_i-\alpha_j}\, . $$ Applying that again and the derivative of $1/x$ you get $$ \frac{\partial^2}{\partial \alpha_i^2} V\, =\, \sum_{\substack{j,k\\|\{i,j,k\}|=3}} \frac{V}{(\alpha_i-\alpha_j)(\alpha_i-\alpha_k)}\, . $$ Then see his Exercise 25 and the simple algebraic fact that facilitates it. Of course this might not be the deeper reason you want. And then you have all the other deeper connections mentioned by the other folks who already gave you answers.

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