These prizes were only recently established and are awarded by/at the International Congress of Basic Science. Interestingly , neither this congress nor the prizes have a wikipedia page yet

• The Breakthrough Prize in Mathematics 2025 goes to Dennis Gaitsgory for his central role in the proof of the geometric Langlands conjecture.
• The New Horizons in Mathematics Prizes go to Ewain Gwynne for his work in conformal probability, John Pardon for producing a number of important results in the field of symplectic geometry and pseudo-holomorphic curves, and to Sam Raskin who has played a significant role in the major recent progress on the geometric Langlands program.
• The Maryam Mirzakhani New Frontiers Prizes go to Si Ying Lee who found a new approach to an important problem in the Langlands program, Rajula Srivastava whos work focuses on bounding the number of lattice points one can find near a given smooth surface, with important applications to Diophantine approximation in higher dimensions, and Ewin Tang who invented quantum computing algorithms for machine learning.

Interestingly, René Thom was able to show that 1) this is not always the case and 2) it is actually the case if we only want to represent in such a way some multiple \(kz\) of \(z\).

If a homology class \(z\) is representable in such a way, the next question one wants to investigate is whether one can improve the map \(f\) in the representation; concretely, can we assume that \(f\) is an immersion or even an embedding? (For this we now assume that \(X\) itself is a manifold.)

A recent paper by Diarmuid Crowley and Mark Grant (arXiv:2412.15359) gives concrete examples that this is not always possible, i.e. an example where \(f\) can not be made an immersion and an example where it can be made an immersion but not an embedding.

It is interesting that despite this being a classical topic, there were no such examples up to now in the literature.

Why did they prove such a result? They wanted to give yet another proof that the only parallelizable spheres are the ones in dimensions 1, 3 and 7 (at that time four other proof already existed). The result that 9-fold suspensions are W-trivial implies, in combination with an argument by Milnor, that spheres of dimensions 9 and higher are not parallelizable.

How did they prove it? They used Bott periodicity for KO-theory. Note first that the computation of the total Stiefel-Whitney class of a real vector bundle \(\eta\) over \(S^9 \wedge Y\) factors through \(KO(S^9 \wedge Y)\). By Bott periodicity we know that this KO-element comes from an element in \(\widetilde{KO}(S^1 \wedge Y)\); concretely, \(\eta\, – k = \beta(x)\), where \(\beta\) is the Bott periodicity isomorphism and k the rank of \(\eta\). The main technical work of Atiyah and Hirzebruch is now to derive a formula for the total Stiefel-Whitney class \(\omega(\beta(x))\) of \(\beta(x)\) in terms of the Stiefel-Whitney classes of \(x\). This formula shows that \(\omega(\beta(x))\) mainly consists of cup products of the Stiefel-Whitney classes of \(x\). But we know that the cohomology ring of \(S^1 \wedge Y\) is trivial (this is true for any suspension) and hence they can conclude that \(\omega(\beta(x))\) is trivial.

Dimension 4 is special in the theory of psc-metrics since here one can use Seiberg-Witten theory to find obstructions to the existence of psc-metrics on closed manifolds which don’t work in higher dimensions. But dimension 4 is also special since here, due to the failure of the h-cobordism theorem, one encounters frequently(*) manifolds with exotic smooth structures, i.e. manifolds which are homeomorphic but not diffeomorphic to each other. Now interestingly, by a new result by Kumar-Sen (arXiv:2501.01113) these two phenomena are closely related: They proved that, at least in the simply-connected case, the \(S^1\)-stability conjecture is true if we allow ourselves to change the smooth structure.

(*) I would like to say that it happens “more often” than in higher dimensions that a 4-manifold admits exotic structures, but I don’t know if this can be made precise of if it is actually true at all.

In a MathOverflow discussion titled Not especially famous, long-open problems which anyone can understand, the moving sofa problem ranks second on the list. If you’re unfamiliar with it, you can find a detailed explanation on its Wikipedia page. Thanks to its intuitive nature, the problem has captured the interest of both professional mathematicians and science enthusiasts alike, leading to coverage in mainstream science media. For instance, Spektrum.de recently featured an article discussing the new breakthrough.

One direction of the conjecture is trivial: If \(M\) admits a psc-metric, then the product metric on \(M \times S^1\) will also have psc. The other direction is the highly non-trivial one and is in its full generality still open, at least in dimensions at least 5 (in dimension 4 one can construct a counter-example using Seiberg-Witten theory). There are two ways to attack the converse direction:

• If \(M\) does not admit a psc-metric, then one tries to show that \(M \times S^1\) does also not admit a psc-metric. This works well if there is a “reason” for \(M\) to not admit a psc-metric, for example based on index theory (in the spin case) or on the minimal hypersurface method (in low dimensions). But if \(M\) only “accidentally” does not admit a psc-metric, it is not clear at all why \(M \times S^1\) should also not admit one.
• The other way, which is of course logically equivalent to the previous one, is to construct a psc-metric on \(M\) given one on \(M \times S^1\). The problem with this approach is that it is hard to construct psc-metrics! But nevertheless this approach was now successfully implemented by Steven Rosenberg and Jie Xu: They proved that if \(M\) is an oriented closed manifold such that \(kM\) (the \(k\)-fold disjoint union) is an oriented boundary, then a psc-metric on \(M\times S^1\) can be used to construct a psc-metric on \(M\) (arXiv:2412.12479).

Note that the oriented bordism ring is very well understood. Especially, the condition that \(kM\) is an oriented boundary for some natural number \(k\) is always satisfied if the dimension of \(M\) is not divisible by \(4\).

edit: As always with fresh preprints on the arXiv one should take these results with a grain of salt …

1. Atiyah-Singer operator: Its topological index is the \(\hat{A}\)-genus and its analytical index can be related to scalar curvature. This shows that the \(\hat{A}\)-genus of a manifold is an obstruction to the existence of a Riemannian metric of positive scalar curvature on it.
2. Signature operator: Its topological index is the \(L\)-genus and its analytical index is the signature of the manifold. This recovers Hirzebruchs signature theorem as a special case of the Atiyah-Singer index theorem.
3. Dolbeault operator: Its analytical index is the holomorphic Euler characteristic and comparison with the topological index recovers the Hirzebruch-Riemann-Roch theorem.
4. Euler characteristic operator: Its analytical index is the Euler characteristic and comparison with the topological index recovers in dimension 2 the Gauss-Bonnet theorem.

Now I personally enjoy the K-homological approach to the Atiyah-Singer index theorem a lot. The main feature of this approach is that elliptic differential operators, i.e. the ones for which Atiyah-Singer applies, define classes in it. So each of the above four operators defines a class in a suitable K-homology group, and one can now investigate this class.

• In the first three cases it turns out that the class is highly non-trivial. In fact, one can show that each of them is an orientation class (in a suitably interpreted sense). For example, for the Atiyah-Singer operator this implies that cap product by it is an isomorphism \(K^*(M) \to K_{m-*}(M)\).
• But in the fourth case, i.e. for the Euler characteristic operator, it is exactly the other way round: It is as trivial as it can be! This was proven by Jonathan Rosenberg (arXiv:math/9806073) and the concrete statement is that the K-homological class of the Euler characteristic operator is in the image of the map \(K_0(\mathrm{pt}) \to K_0(M)\) induced by the inclusion of a point.