## Hot answers tagged 4-manifolds

57

The usual convolution method for approximating continuous maps by smooth maps does not succeed in approximating invertible [resp. injective] continuous maps by invertible [resp. injective] smooth maps.
(referring now to the "new wrong-headed attempt" in a comment at Neil's thread) A sequence of diffeomorphisms or smooth embeddings may converge uniformly to ...

42

Let's try doing this with a compact, closed, connected $1$-manifold $M$. Certainly I can choose a topological embedding $f:M\to\mathbb{R}^2$. However, the image might be very fractal, like a Koch snowflake for example. If I try to take a short piece of the image and straighten it out by projecting in some direction, the fractalness will ensure that I lose ...

33

$\mathbb RP^3$ double-covers the lens space $L_{4,1}$, so it's the boundary of the mapping cylinder of that covering map.
In general $\mathbb RP^n$ for $n$ odd double-covers such a lens space. So in general $\mathbb RP^n$ is the boundary of a pretty standard $I$-bundle over the appropriate lens space. To be specific, define the general $L_{4,1}$ as ...

32

To play devil's advocate, you could easily turn around this line of thought. Since we live in a 3+1-dimensional space-time, we develop concepts that are sensitive to our experiences. So calculus and manifold theory single out dimensions 3 and 4 because that's the only way we know how to design geometric things. Of course the definitions of smooth ...

24

A quick answer...
Likely the most well-known and accepted paper in the physics community on this is, no surprise, by Witten Global gravitational anomalies.
There he argues, in dimensions higher than 4, that an exotic differential structure should be interpreted as a gravitational instanton. He punts on the four dimensional case as the generalized Poincaré ...

22

This is a notorious open problem. For the moment the simplest compact four-manifold that is announced to admit (infinite number of) exotic smooth structures is $S^2\times S^2$. This result is contained here : http://arxiv.org/abs/1005.3346
I have to say that I am not at all an expert in the area
(also it seems that the above paper is not yet published). On ...

22

$\mathbb{RP}^3$ is the unit (co)tangent bundle to $S^2$. Thus it bounds the disc bundle in $TS^2$. Alternatively, any time you have a 2-sphere of self-intersection $\pm 2$ in a closed 4-manifold - for example, any Lagrangian sphere in a symplectic 4-manifold - you get a splitting into two pieces along an $\mathbb{RP}^3$. So, you could consider the diagonal ...

20

I would also like to know the answer to that. As far as I know, it is still a difficult, unsolved problem. The bit about 3-handles is a clue, but I haven't found any way to make use of it.

19

Alexander's horned sphere is a topological sphere in 3-space that cannot be "ironed out", otherwise we would get a smooth (or PL) 2-sphere having a complementary region which is not simply-connected, a fact which is excluded because every smooth (or PL) 2-sphere in 3-space is standard.
Another simple topological object that cannot be smoothly ironed is a ...

18

Thanks to Ian Agol for pointing out this question and a related one on levels of Morse functions - /305067/. In both case, for smooth manifold of dim $> 3$, as expected, there is no finite list of blocks ( or regular level components.) The idea is that one may define the "width" of a group, by representing the group G as the fundamental group of some ...

18

I can answer your literal question. Not everyone studies exotic $\mathbb{R}^4$, because the universe of mathematical and theoretical physics is a big one with many interesting ideas, and there's no reason for everyone to drop all of their potentially fruitful projects to study one speculative idea that may be a dead end. Even though we have a continuum of ...

18

The 4-dimensional geometries were classified in the unpublished thesis of Filipkiewicz, which is available here.

17

There are not that many explicit handlebody pictures of exotic open 4-manifolds, because they get awfully complex in short order. The ones that I know of are in work of Žarko Bižaca from the mid-90's. I think you probably can work out this particular case by hand. You don't really want to use Quinn's theorem for this, because it is not exactly ...

17

If an $n$-dimensional smooth manifold $X'$ is obtained from $X$ by doing some surgeries, then there is an $(n+1)$-dimensional smooth cobordism $W$ from $X$ to $X'$; vice-versa, any handle decomposition of such a cobordism induces a sequence of surgeries.
Hence, the equivalence relation of "being obtained from one another by surgeries" is equivalent to the ...

16

I am not sure how relevant it is, but here is something that made me better understand the possible problems when one tries to smooth out an embedded topological manifold.
Take your favorite knot $K$ in $S^3$ (or, if you are minimalist, take your second favorite knot so that it is not trivial), view $S^3$ as an equator in $S^4$, and consider the cone over ...

15

This is a well-known open problem. In fact, there are very few tools for studing general negatively curved manifolds. Even in dimension 3 it is unknown (I think) how to prove existence of proper finite index subgroups without using the geometrization. Geometrization implies residual finiteness of f.g. 3-manifold groups, and hence existence of proper finite ...

15

One basic structural problem about the SW invariants is the question of simple type: suppose that $X$ is a simply connected 4-manifold with $b^+>1$, and $\mathfrak{s}$ a $\mathrm{Spin}^c$-structure such that $SW_X(\mathfrak{s})\neq 0$. Must $\mathfrak{s}$ arise from an almost complex structure? This is true when $X$ is symplectic (Taubes in ...

14

This can be answered by obstruction theory for the fibration
$$ F=SO(4)/U(2) \to BU(2) \to BSO(4) $$
where the fibre is actually a 2-sphere: $F=S^2$. Start with the tangent bundle of an oriented 4-manifold $M$ and ask for existence respectively uniqueness of a lift of its Gauss-map $M\to BSO(4)$, that is, existence respectively uniqueness for an almost ...

14

The conjecture that every $S^2 \subseteq \mathbb{C}P^2$ is standard if it is homologous to flat is implied by the smooth Poincaré conjecture in 4 dimensions. It also implies a special of smooth Poincaré that is accepted as an open problem, the case of Gluck surgery in $S^4$. I can't prove or disprove the question of course, but since the question is ...

14

In algebraic geometry, there are examples of "fake projective planes", which in this context means smooth complex surfaces of general type with the same cohomology ring as the complex projective plane. It is known that the universal cover of such spaces is the complex hyperbolic ball. So the answer to your question is yes. (The first such fake projetive ...

13

I have to apologize, in fact the answer to the second question is still unknown. Namely, up to now all known symplectic manifolds of dimension 4 that have negative Euler characteristic are blow ups of ruled surfaces. However it is not known if there are no other examples. I have corrected the answer accordingly.
It is difficult to give a precise answer to ...

12

This was conjectured by Witten in his paper in his paper Monopoles and four-manifolds. The conjecture says that if $X$ has Seiberg-Witten simple type (meaning that $SW_X(\mathfrak{s})$ is nonzero only when the moduli space associated to $X$ is 0-dimensional) and satisfies some mild homological conditions (including $b_1(X)=0$ and $b^+(X)>1$ odd), then ...

12

There is a long, long list of mathematical subjects that were either pioneered or significantly inspired by results in quantum field theory. However, while physicists may trust the manipulations they do in QFT, and the results of those manipulations have been spectacularly successful, for almost every interesting quantum field theory, there isn't even a ...

12

As stated, your question is equivalent to the existence of a large exotic 4-ball (a smooth $D^4$ which cannot be smoothly embedded into $\mathbb{R}^4_{std}$).
The existence of a flip would give rise to an exotic $S^4$, by gluing the $D^4$ at infinity using the flip diffeomorphism. Removing a (small) standard ball from this $S^4$ gives a large exotic $D^4$, ...

12

No. Suppose that the rank of $H^1(V_1)$ is zero, so that the rank of $H^1(V_2)$ is three and (by looking at the Mayer-Vietoris sequence again) the rank of $H^2(V_2)$ is zero. Take two independent elements of $H^1(V_2)$. Their product in $H^2(V_2)=0$ is trivial, but its image in $H^2(M)$ is nontrivial, being the product of two independent elements of ...

12

I wrote this up for my own personal edification when I was going through Gompf and Stipsicz. Please, excuse the pedantry. I had to make sure even I understood it.
Definition: (Smoothly Slice Knot)
A knot $K$ in $\partial D^4$, where $D^n$ is the standard closed $n$--dimensional disk, is smoothly slice if there exists a two-disk $D^2$ smoothly embedded in ...

11

Related to the early investigation of the Thom conjecture,the G-signature thm was used circa 1970 to give 4-ball genus bounds for torus knots which asyptocically (in some cases) were a fixed fraction of what we now know to be the smooth category answer. I belive Larry Tayor observed (in the '70s or early 80s) that these G-signature bounds hold in the ...

11

One basic problem is determining the relationship between Seiberg-Witten invariants and Donaldson invariants of $4$-manifolds. Witten himself proposed the precise relationship between the two in the original paper Monopoles and 4-Manifolds, but as far as I know the relationship has not been proven in general. Witten's conjecture has been proven in many ...

11

A biased answer, based on Auroux's work http://arxiv.org/abs/1003.2962.
Auroux makes a connection between bordered Floer theory and an alternative approach, due to Lekili and myself, which is (still) under development, but which should include the $\pm$ and $\infty$ versions. We do have a preliminary paper out: http://arxiv.org/abs/1102.3160.
A general ...

11

By Poincaré duality, there is an isomorphism
$H_2(M, \mathbb{Z}) \cong H^2(M, \mathbb{Z})$.
Now let $PD(a) \in H^2(M, \mathbb{Z})$ be the Poincaré dual of $a$. Since $H^2(M, \mathbb{Z})$ classifies line bundles on $M$, there exists a line bundle $L$ such that $c_1(L)=PD(a)$. Take a general smooth section of $L$. Then its zero set is a smoothly embedded ...

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