@BalarkaSen yeah, this is actually clear. if you take an $m$-sphere with the canonical framing in a coordinate ball in $M$, then surgering $M$ along that sphere yields the connected sum of $M$ and the analogous surgering of $S^n$ along a canonically framed $m$-sphere. however, decomposing $S^n=S^m\times D^{n-m}\cup D^{m+1}\times S^{n-m+1}$ in $D^{n+1}=D^{m+1}\times D^{n-m}$, we clearly see that this is $S^{m+1}\times S^{n-m-1}$.
'actually clear' meaning I stared blankly at a piece of paper for 30 minutes till I realized it was actually clear
god I've been sucking at surgery these past 2 days
@BalarkaSen I think (2) may be free? Alexander-Lefschetz duality identifies these with the last two terms in the exact sequence $H^1(W)\rightarrow H^1(N)\rightarrow H^2(W,N)\rightarrow H^2(W)$, but the first map is an isomorphism since $N\rightarrow W$ is an iso on $\pi_1$, right?
anyway, this sounds cool!
I'm pretty sure I heard Matthias talk about this stabilization with $S^2\times S^2$ stuff before, though I don't remember specifics
@BenSteffan nobody knows what's a manifold is, that's why we invented surgery theory to describe the abstract (manifolds) in terms of the concrete (CW-types)
@Thorgott dug up my old lecture notes again and I'm reading that the set of $4$-manifolds with a fixed fundamental group $\pi$ up to stable diffeomorphism (i.e. after $\#$-ing with $S^2\times S^2$'s) is classified by $\mathbb{Z}\times H_4(K(\pi,1),\mathbb{Z}^{w_1})$...
that was so beyond me
@BenSteffan I've had to explain unironically that a lecture wasn't about four specific manifolds too many times before
it's not a complete classification, but these guys have an invariant living in $\mathbb{Z}\times H_4(K(\pi,1),\mathbb{Z}^{w_1})$ and the Theorem was that every value of this invariant gets realized
@BenSteffan imagine my disappointment when I first learned they aren't pronounced like that
A professor once told us on a lecture that there are many ways to deal with partial fraction and there are many books written about this, I don't know if he was exaggerating or not (it is quite hard to tell with him since he like to exaggerate a lot) but are there really books on dealing with partial fractions?
@Thorgott Right, that's it. Do it in a ball and pop the ball out to get a connect sum decomposition.
@Thorgott This is confusing me a bit. For example, take the cobordism $W : M \Longrightarrow N = M \# (S^2 \times S^2)$ you constructed. Then $H_2(W, M \sqcup N; \Bbb ZG) = 0$ but $H_2(W, M; \Bbb ZG) = \Bbb ZG$, is it not?
Ah you have a dimension error. $\dim W = 5$, so Alexander-Lefschetz duality identifies $H_2$ with $H^3$.
However, a corrected version of your Alexander-Lefschetz duality and exact sequence argument shows if $H^2(W; \Bbb ZG) \to H^2(N; \Bbb ZG)$ is surjective, then we are through. This is the same as saying $\pi_2(N) \to \pi_2(W)$ is injective (again, we use $G$ is finite).
@Thorgott I learnt that this is the general strategy in Kreck's modified surgery (from his 1999 Annals paper): his techniques classify $2q$ manifolds (with $B$-structures) upto stabilization. Then one would like to cancel $(S^q \times S^q)$'s somehow.
@Thorgott I think if the manifold has universal cover non-spin, then it is indeed a complete set of invariants for a stable diffeomorphism classification.
Here's what I was thinking for (2). Suppose $W : M \Longrightarrow N$ is a cobordism with $M, N \hookrightarrow W$ isomorhism in $\pi_1$, and $H_2(W, M \sqcup N; \Bbb ZG) = 0$. If $H_2(W, M; \Bbb ZG) \neq 0$, then choose an embedded sphere $S \subset W$ such that $[S] \neq 0 \in H_2(W, M; \Bbb ZG)$ is a non-zero class. We know $[S] = 0$ rel $M \sqcup N$, so $[S]$ must be in the image of $H_2(N; \Bbb ZG) \to H_2(W; \Bbb ZG)$.
Therefore, $[S]$ is in the image of $\pi_2(N) \to \pi_2(W)$, so $S$ is homotopic to a $2$-sphere in $N$.
If $[S]$ is not $\Bbb Z$-torsion, the universal coefficients dual of $\mathrm{PD}[S] \in H^3(W, N; \Bbb ZG) \cong H^3(\widetilde{W}, \widetilde{N}; \Bbb Z)$ (using $G$ finite to get rid of compact support as $\widetilde{W}$ is compact, once again) must be a non-zero class in $H_3(\widetilde{W}, \widetilde{N}; \Bbb Z)$. Therefore, by a similar argument, the meridian $2$-sphere $T$ of $S$ must be homologous in $\widetilde{W}$ to $\widetilde{N}$.
So $[T]$ is in the image of $H_2(\widetilde{N}; \Bbb Z) \to H_2(\widetilde{W}; \Bbb Z)$, i.e., in the image of $\pi_2(N) \to \pi_2(W)$, as well. So if $j : N \to W$ is the inclusion map, there exists $\alpha, \beta \in \pi_2(N)$ such that $j_\star \alpha = [S]$, $j_* \beta = [T]$. Since $S$ and $T$ link once in $W$, $\alpha$ and $\beta$ must have intersection number $+1$ in $N$.
For any manifold $N$ with fundamental group $G$, $\pi_2(N) = H_2(N; \Bbb ZG)$ carries a $\Bbb ZG$-valued equivariant intersection form. I think the above tells me that the subspace spanned by $\{\alpha, \beta\}$ is the quadratic space $H \otimes \Bbb ZG$, where $$H = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$ is the hyperbolic form.
Although I have not checked $\alpha, \beta$ self-intersect trivially in $N$. But that should be true.
I am from the physics side. I would like help with the mathematics side of an exercise I am dealing with. Would it be possible to discuss something like that here?
Hi guys I am from the physics side. I would like help with the mathematics side of an exercise I am dealing with. Would it be possible to discuss something like that here?
If we are given ax + by = 1 (everything is an integer), then obviously a(kx) +b(ky) = k, now if we are given another equation ax' + by' = k, then is it necessary that x' and y' are multiples of k?
Everything is an integer in the second equation too.
@Thorgott I think not necessarily! Perhaps you have a pair of immersed $2$-spheres in $N$ which have trivial normal bundle each, and they intersect transversely once.
Since we are in dimension $4$, we cannot make these spheres into an embedding.
For instance, take let's say the $K3$ surface. This is simply connected, and has intersection form $2E8 \oplus 3H$.
However, $K3$ is not of the form $M \# (S^2 \times S^2)$ for some other $4$-manifold $M$.
But if the arguments above go through, one (perhaps) gets the following: For a finite group $G$ such that stably free $\Bbb ZG$-modules are free and $Wh(\Bbb ZG) = 0$, if $M$ and $N$ are two 4-manifolds with fundamental group $G$, the $\Bbb ZG$-valued equivariant intersection form of $N$ does not split off an $H \otimes_{\Bbb Z} \Bbb ZG$, then $M$ and $N$ are stably diffeomorphic implies $M \cong_{top} N$.
The two hypothesis go through for, for instance, $G = S_n$, $n \geq 5$. Symmetric groups have trivial Whitehead torsion, and stably free modules over them are indeed free.
@Thorgott The following is a result of Hambleton & Kreck. Let $M, N$ be $4$-manifolds with finite fundamental group $G$, and suppose $M \#^k (S^2 \times S^2) \cong_{\mathrm{diff}} N \#^k (S^2 \times S^2)$. If $M = M_0 \# (S^2 \times S^2)$, then $M \cong_{\mathrm{top}} N$.
There they require a splitting off of an $H$ (space-level, so a genuine $\# (S^2 \times S^2)$), whereas I guess I am requiring the opposite condition.
Correction: $K3$ is not of the form $M \# (S^2 \times S^2)$ for some other smooth $4$-manifold $M$. I suppose $\#^2 E8 \#^2 (S^2 \times S^2)$ works just fine, by Freedman's theorem.
I agree, but maybe I am more worried about "$N$ does not split off an $H \otimes_{\Bbb Z} \Bbb ZG$" -- can we cook up examples of such $4$-manifolds (ideally, smooth?).
There's a full classification of finite groups $G$ for which stably free $\Bbb ZG$-modules are free.
It is due to Jacobinski. Theorem: $G$ be a finite group which has no quotients isomorphic to the generalized quaternion group $Q_{4n}$, $n \geq 2$, binary icosahedral group, binary octahedral group or the binary tetrahedral group. Then stably free $\Bbb ZG$-modules are free.
$Wh(\Bbb ZG) = 0$ looks more annoying.
But maybe that can be improved, because of this remark I'd made earlier.
@BalarkaSen Of course, $S^4/\Gamma$ are examples. But I have a feeling the standard construction $X(G)$ of a $4$-manifold with given fundamental group $G$ also has this property.
I learnt the symmetric groups have Wh = 0 from Milnor's article. Perhaps more is known now.
It must be possible to relax it quite a bit, because I do not really want Wh = 0 but a specific element in Wh is zero.
@Thorgott Here's another ridiculous question. Given an element $[A] \in Wh(\Bbb ZG)$, one gets elements in $Wh(\Bbb F_p G)$ and $Wh(\Bbb Q G)$ by reducing the matrix $A$ modulo $p$ or treating it as a matrix over the rationals. Are vanishing of the latter sufficient for vanishing of the integral element?
In general, is there a local-global principle for $K_1$
@Thorgott The proof is surprisingly neat! Jacobinski studies instead the group ring $\Bbb RG$, which decomposes into division algebras over $\Bbb R$ by Wedderburn since $G$ is finite. He points out that the only problematic division algebra, which gives rise to stably free non-free modules over $\Bbb ZG$, are quaternions.
Those groups are the only ones for which the left-regular representation of $G$ has a quaternion component in its decomposition into simple modules.
There's a converse due to Swan which rules out some of the groups from the "bad list". But not having anything from there is enough.
the vanishing mod $p$ tells you something due to exact sequence of $K_1$'s, but I don't see how to really meaningfully leverage that
also, I checked a reference, and $Wh(\mathbb{Z}G)$ is infinite for most abelian $G$ (any $G$ containing an element of order $5$ or $\ge 7$, for example)
@Z.G. I suppose so. ;) It's probably easiest to start with the basic form where gcd(a, b) = 1, so ax + by = 1 (Bezout's identity) has solutions. en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
To be honest, I don't remember the proof off-hand that $a(x+bt) + b(y-at) = k$ gives all solutions. I'm sure I proved it several decades ago, though. ;)
I'm half asleep, and don't feel like proving it right now. Sorry.
> Every closed subset $E$ of a complete metric space $X$ is complete.
Proof sketch in Rudin's book: every Cauchy sequence in $E$ is a Cauchy sequence in $X$, hence it converges to some $p\in X$, and $p\in E$ as $E$ is closed.
I'm trying to justify this last part (i.e. $p\in E$) by the theorems proved earlier in the book. For instance, there's a theorem (Theorem 3.2) saying if $E\subset X$ and $p$ is a limit point of $E$, then there's a sequence in $E$ converging to $p$. It seems like I need the converse of this statement, but the converse is false. I feel stuck.
the full set of solution is given by $a(x+bt/k)+b(y-at/k)$ as $t$ ranges over the integers
cause any two solutions differ by a solution to $ax+by=0$, as I said, an obvious solution to which is $(b,-a)$. the solutions are all real multiples of another and the smallest integer multiple of this is $(b/k,-a/k)$.
@PM2Ring The case where GCD = 1 is the one I had in mind (I was proving that two consecutive Fibonacci numbers are coprime using Bezout's lemma, that's horrible but I managed to prove it simply by contradiction)
@Thorgott that the solutions differ by solutions to ax+by=0
@Z.G. Wikipedia mentions an even stronger result: With the exceptions of 1, 8 and 144 every Fibonacci number has a prime factor that is not a factor of any smaller Fibonacci number.
FWIW, a year or so ago I was messing around with efficient algorithms for calculating insanely large Fibonacci numbers. These algorithms are of limited practical use because the numbers are so large. :)
@Z.G. Binet's formula is ok, but for really big Fibonacci numbers you need to know phi to a lot of decimal places. And it uses real number arithmetic. It's more efficient to use integer arithmetic.
@Thorgott this is true irrespective of the claim in Atiyah-Bott, I've proven it by hand (corrections: assume $X$ paracompact or all bundles numerable, also the homotopy classes in the latter category should be both equivariant and fiberwise)
At the bottom of the box of shredded wheat squares, there's quite a bit of 'loose' shredded wheat; it used to be in cereal form and is now just loose.
What interesting recipes or dish modifications can you suggest?
One idea was a sort of crisp/struedel topping, (mix with butter and bake on top ...
i dunno about that. if you grind it fine enough you could increase the quantity of something else with it. turn a kilo of cocaine into 1.2 kilo of 'cocaine'
Not because it collapsed the cocaine economy, but because there was just so much damn cocaine around. Why do you think they started putting it in Coke?
In Wikipedia page en.wikipedia.org/wiki/Binary_icosahedral_group it is stated that the binary icosahedral group is the smallest superperfect group[citation needed]. I am not able to find a proof of this statement. Can someone provide a proof or a reference to a proof?
At the bottom of the box of shredded wheat squares, there's quite a bit of 'loose' shredded wheat; it used to be in cereal form and is now just loose.
What interesting recipes or dish modifications can you suggest?
One idea was a sort of crisp/struedel topping, (mix with butter and bake on top ...
