I know this question sounds like a joke. And it probably is:).
I found it kind of annoying, but also interesting, to call $-\Delta=-\sum_{j=1}^n\partial^2_{jj}$ "the positive Laplacian" as it is the positive operator with respect to standard $L^2$-pairing. But in the mean time it is also regarded as the "negative of Laplacian" since we have a minus sign.
I think there are two different questions related to this topic.
- What is a more clear way to explain the difference on what positive and negative is for a Laplacian? And also how to use such terminologies appropriately in different circumstances.
I believe such clarification is important, as the Hodge-Laplacian $\Box=dd^*+d^*d$ (let me not using $\Delta$ here) can be called a "positive Laplacian" with slightly less ambiguity. Also the word "subharmonic functions" refers to those $f$ such that $-\Delta f\le0$, but not $\Delta f\le0$.
- Are other similar (but also interesting) confusions happen in other fields of mathematics? If I remember correctly, the meaning of covariant tensors and contravariant tensors in commutative algebra and in differential geometry are flip. They seem less "annoying" than Laplacian though.
We call $-\Delta$ the positive Laplacian since it is positive semi-definite as a self-adjoint operator on, say, $C_c^\infty({\mathbb R}^n)$, the space of functions with compact support under the $L^2$ inner product. That is, if $f, g\in C_c^\infty({\mathbb R}^n)$, then \begin{align*} \langle -\Delta f, g\rangle &= \int_{{\mathbb R}^n} (-\Delta f) g = \int_{{\mathbb R}^n} f(-\Delta g) = \langle f, -\Delta g\rangle,\\ \langle -\Delta f, f\rangle &= \int_{{\mathbb R}^n} (-\Delta f) f = \int_{{\mathbb R}^n} |\nabla f|^2 \geq 0, \end{align*} where the identities hold by integration by parts and there are no boundary terms by the compact support condition. Then the eigenvalues of $-\Delta$ are nonnegative.
The Hodge-Laplacian $\square = dd^* + d^*d$ is, by construction, self-adjoint and positive semi-definite. Actually on a function $f$ on a compact manifold, $$ \square f = - \Delta f. $$
A $C^2$ function is subharmonic iff $\Delta f\geq 0$, and the reason is that such function lies below a harmonic function with the same boundary condition.