Let $M$ be a simply connected oriented closed 4-manifold. Then $H^2(M)$ is a free abelian group, with a pairing $H^2(M) \otimes H^2(M) \to H^4(M) \cong \Bbb Z$ (the last isomorphism coming from our chosen orientation). The bilinear form this determines on $H^2(M)$ is called the intersection form. It has determinant 1 by Poincare duality. We say it's 'positive definite' if $a \cup a \geq 0$ for all $a$. Freedman proved that every unimodular positive definite intersection form is realized
by some s.c. closed oriented topological 4-manifold, and is mostly unique up to homeomorphism. (To be precise: It's unique up to homeomorphism if $a^2$ is always even, and there are exactly two if $a^2$ is odd; at most one of the two will be smoothable.)
But his construction only works in the topological category. It says nothing about smooth 4-manifolds. Donaldson, under the advice of Atiyah, was doing something completely different. If $M$ above is smooth and has a Riemannian metric, you can write down the space of connections on an $SU(2)$-bundle over it "with charge $k$" (let's just say $k$ measures how twisty the bundle is). There's a PDE on this space, called the ASD equation: $F_A^+ = 0$.
If $\mathcal C$ is the space of connections, there's a (n infinite dimensional) group $\mathcal G$ acting on it, that preserves the set of points with $F_A^+ = 0$. That is, if $F_A^+ = 0$, then $F_{g(A)}^+ = 0$ as well. So I can consider the "moduli space of solutions", $\mathcal I/\mathcal G$, where $\mathcal I$ is the set of points with $F_A^+ = 0$.
This moduli space is not compact, but it has a natural compactification (called the "Uhlenbeck compactification") involving adding solutions from bundles with smaller instanton number. Donaldson proved the following: If $M$ is positive definite, then the compactified moduli space $\mathcal M_1(M,g)$ of instantons on the bundle with instanton number 1 is almost a smooth manifold for generic $g$ (but not for all $g$)
Away from a finite set of points, it has the natural structure of a smooth manifold with boundary, and that boundary is $M$ itself. The finite number of singularities where it's not a smooth manifold are locally homeomorphic to the cone $C(\Bbb{CP}^2)$.
Indeed, he can write down explicitly how many there are: there are precisely as many as there are elements in $H^2(M)$ with $a^2 = 1$
Deleting a small neighborhood of the cone points you get a cobordism from $M$ to that many copies of $\Bbb{CP}^2$; because cobordisms preserve signature, there must be precisely $b_2(M)$ of them. You can use this data to explicitly prove that the intersection form is diagonalizable over the integers.
Thus, every smooth positive definite etc 4-manifold has diagonalizable intersection form. But Freedman produced a topological manifold for every positive definite bilinear form. There are a lot of those; the number of positive definite unimodular bilinear forms grows exponentially in the rank of $H^2(M)$.
So Donaldson's theorem and Freedman's theorems combine to prove that there are many, many non-smoothable 4-manifolds.
Donaldson's theorem is the flavor of the kind of math I think a lot about.
oh, does it also matter if the permutation acts on a 3 term polynomial in the form $ax_1x_3x_2 - bx_2x_4 + cx_1x_2x_3x_4$? (with varying exponents) I'm focused on the last term mostly
(Actually Freedman's theorem already provided many; I told you that when the intersection form is even he gets two, one which is not smoothable. But you can measure the non-smoothability of this second one (actually non-PLability, but whatever) from an invariant called the Kirby-Siebenmann invariant which is somehow a more classical obstruction to smoothability, so it's not that surprising.)
There's a famous conjecture called the 11/8 conjecture which more or less says that every simply connected smooth 4-manifold is homeomorphic to one of $\# k\Bbb{CP}^2 \# \ell \overline{\Bbb{CP}^2}$ or $\# k(K3) \# \ell S^2 \times S^2$, where $K3$ is the $K3$ surface, aka zero set of a quartic curve in $\Bbb{CP}^3$. But this conjecture seems very out of reach at present.
@arctic unless this is a different type of permutation than what I'm used to, I expect $(1~2~3~4)$ to permute $x_1^6$ to $x_2$ , $x_2 \to x_3$, etc. to get $x_4^6 x_1 x_2 x_3^{23}$
In many of those manifolds we know that there are infinitely many non-diffeomorphic smooth structures. These frequently are distinguished by doing something along the lines of "counting the number of points in $\mathcal M_0$", which is a smoothish invariant.
@Danu Freedman's theorem says there aren't that many simply connected smooth 4-manifolds. The simply connected assumption is trivial. In general $(n/2-1)$-connected manifolds are classifiable. $n= 3, 4$ are the odd ducks.
@Semiclassical I have a bunch of interesting stuff that I put from other papers on a single notebook under the words Very nice and special :-), but some I didn't write correctly.
@ForeverMozart Be careful that pride in mathematics is very dangerous. I mean you can be proud, boastful in front of anybody, but not to yourself, be honest with yourself, otherwise you go down with very high speed. :-)
@ForeverMozart I didn't mean anything bad with that, but just to be careful. There is always room to learn from others, and this is a (very) good thing. I admit that for a long period of time I develop only my research ideas, but at the same time I know that at some point I need to return and also consider other new ideas from others.
If $a,b,c$ are positive real numbers, show that $$\frac{1}{a+\sqrt{bc}}+\frac{1}{b+\sqrt{ac}}+\frac{1}{c+\sqrt{ab}}\le \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
Or this one (which is again very cool)
(typing)
Let $a,b,c>0$ such that ab+bc+ca=1. Show that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\sqrt{3}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}$$
