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5h
comment How to recover multiplication of group elements from category of groups?
@Zhen: sure, but I'm trying to address the motivating question. My claim is that the Lawvere theory of groups already constitutes "everything about groups." I agree that there's still an interesting additional question remaining.
6h
answered How to recover multiplication of group elements from category of groups?
6h
comment Why is this not an inconsistency in elementary Lie theory?
It's for the same reason that in order to translate the graph of $y = f(x)$ $c$ units to the right, you have to subtract $c$ to get $y = f(x - c)$.
8h
comment Intersection preserves homotopy equivalence
I think the first $X$ was supposed to be a different letter.
8h
answered Intersection preserves homotopy equivalence
1d
comment Nth Homotopy Group Isomorphic to [T^n, X]
$S^n$ and $T^n$ are not even homotopy equivalent (except when $s = 1$; they can be distinguished by $H_1$), so there's no hope for them to be equivalent as anything more structured.
1d
comment Are the Spherical harmonics the S^2 equivalent of the exp(i \pi n) function series?
One important difference is that $S^2$ is not, say, a locally compact abelian group, so there isn't, strictly speaking, a Fourier transform on it. (So, for example, one property of the usual Fourier transform is that it intertwines pointwise multiplication and convolution. But what does convolution mean on $S^2$?)
2d
comment Sum of Betti numbers of a degree $d$ hypersurface of $\mathbb{P}^n_{\mathbb{C}}$
It means that all of the Tor and Ext terms in the universal coefficient theorems for both homology and cohomology vanish. You only need to invoke the Lefschetz hyperplane theorem over $\mathbb{Z}$, and after that it's just a matter of using the universal coefficient theorem.
2d
answered raising elements of profinite groups to $p$-adic powers
2d
comment Can one construct any n-gon if angle trisection is also allowed?
It suffices to allow $n$-sections for all prime $n$, but maybe that's obvious. More interesting is that I think if one prime is omitted then it's not enough (but I haven't thought this through carefully).
2d
awarded  Good Answer
2d
comment $K_0$ of a ring via idempotents
Yes, that's correct.
Aug
28
comment Symplectic form and wedge sum
@Mike: no, that's only if we're talking about antisymmetric tensors; the OP is explicitly using the wedge product symbol, so there should be a unique expression giving the $k^{th}$ wedge power.
Aug
27
comment Infinite dimensional representation such that every subrepresentation is reducible
It's not true that if $V$ has no irreducible subrepresentations then it is itself irreducible; see the other answers. And what is the contradiction? Among all sequences, which might have finite or infinite support, there are some that have finite support. I claim this is a subrepresentation (although not an irreducible one); moreover, it is also infinite-dimensional.
Aug
27
comment Infinite dimensional representation such that every subrepresentation is reducible
Ah. That's still not irreducible, though: for example, there is a subrepresentation consisting of the sequences with finite support (so $c_i = 0$ for sufficiently large and sufficiently small $i$). And again, that's still not what the question asked for.
Aug
27
comment Infinite dimensional representation such that every subrepresentation is reducible
What you've written down is a representation of the monoid $\mathbb{Z}_{\ge 0}$, not the group $\mathbb{Z}$ or the ring $\mathbb{Z}$. The ring you want to talk about is, say, $\mathbb{Z}[x]$ (the monoid ring of $\mathbb{Z}_{\ge 0}$). I don't understand what you're trying to claim: this representation very much fails to be irreducible, but that's not what's being asked for anyway.
Aug
27
comment Symplectic form and wedge sum
You mean the wedge product? In any case, I don't see any reason why that $k!$ prefactor should be there.
Aug
27
answered Algebraic integers of $\mathbb{Q}(\sqrt{m})$ for $m$ a squarefree integer
Aug
27
comment $H^{p,0}$ is isomorphic to the space of holomorphic forms?
What exactly is your question?
Aug
27
comment Smallest example of a group that is not isomorphic to a cyclic group, a direct product of cyclic groups or a semi direct product of cyclic groups.
But we need a stronger result, namely that it is not even isomorphic to a semidirect product.