bio | website | |
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location | ||
age | 24 | |
visits | member for | 4 years, 9 months |
seen | 1 hour ago | |
stats | profile views | 11,202 |
Jul 18 |
awarded | Enlightened |
Jul 18 |
awarded | Nice Answer |
Jul 18 |
awarded | Nice Answer |
Jul 18 |
comment |
A question on fixed point theory
I don't entirely follow how you're defining your map: what does the map send $U$ to? What if $U$ is orthogonal to $\mathbb{C}^{k+1}$? How is the unitary you're specifying unique? In any case, it doesn't appear that your argument ever uses that $k>0$. |
Jul 18 |
comment |
A question on fixed point theory
More generally, $n$ needs to be large enough compared to $k$ to ensure the existence of an integer $d$ as in the argument above (I haven't checked carefully, but I think $n\geq (k+1)(2k+1)$ should suffice). The condition that either $k$ is odd or $n$ is even is just so that it is impossible for $1+d^{k+1}+\dots+d^n$ to be zero. |
Jul 18 |
comment |
A question on fixed point theory
Consider, for instance, $\mathbb{C}P^6/\mathbb{C}P^1$. Its cohomology is the subring of $\mathbb{Z}[x]/x^7$ generated by $s=x^2$ and $t=x^3$. We have a relation $s^3=t^2$, and this forces any endomorphism of the ring to be of the form $s\mapsto d^2s$, $t\mapsto d^3 t$ for some $d\in\mathbb{Z}$. Since $1+d^2+d^3+\dots+d^6$ can never be zero, every map has a fixed point by the Lefschetz fixed point theorem. |
Jul 14 |
comment |
A question on fixed point theory
An slight variation of the usual cohomological argument should give an easy positive answer when $n\gg k$ and $k$ is odd or $n$ is even. For instance, for $k=1$ and $n\geq 6$ every endomorphism of the cohomology ring of $\mathbb{C}P^n/\mathbb{C}P^1$ extends to the cohomology ring of $\mathbb{C}P^n$, and so the fixed point property follows by an easy computation. |
Jul 13 |
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Are Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$?
Bounded analytic functions are not "completely regular", though; the idea behind my example was basically to enlarge them to make them completely regular. |
Jul 13 |
revised |
Are Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$?
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Jul 13 |
answered | Are Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$? |
Jul 12 |
reviewed | Leave Open is grassmannian rational connected or not |
Jul 6 |
comment |
Is the monoid of taking iterated images and inverse images freely generated by the image and inverse image operation?
@TheMaskedAvenger: If I'm not mistaken, any bijection will completely fail to separate words (i.e., it will only separate those that are necessarily distinguished for reasons of degree or parity), whereas by my argument any non-bijection will separate all words. |
Jul 6 |
answered | Is the monoid of taking iterated images and inverse images freely generated by the image and inverse image operation? |
Jul 6 |
comment |
Is the homotopy category of a ring also the derived category of another ring?
I expect this is usually false. For instance, $D(S)$ always has a compact generator, and it seems unlikely that $K(R)$ does (except maybe if $R$ is Artinian or some condition of that sort). Indeed, it's not clear to me that $K(\mathbb{Z})$ even has a set of generators, though I don't have very strong intuition for this. |
Jul 5 |
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R[[X]] flat as a R[X]-module?
@user52824: Why does it suffice to consider the case when $R$ is a domain? It is not true that $R/I[[x]]=R[[x]]\otimes_{R[x]} R/I[x]$ if $I$ is not finitely generated. |
Jul 5 |
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R[[X]] flat as a R[X]-module?
@VesselinDimitrov: The localization of a ring at a prime need not be a domain. The ring $R=k[t,y_1,y_2,\dots]/(ty_1,ty_2,\dots)$ is such that $R[[x]]$ is not flat over $R[x]$ (consider the ideal generated by $t$ and an element like $\sum y_n x^n\otimes t$). |
Jul 4 |
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Properties of rings that have an elegant description in terms of the associated category of modules
Surely the ideal $(1)$ can be considered as the empty product of primes. |
Jul 2 |
awarded | Curious |
Jun 27 |
comment |
Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Do you mean to assume that $0\leq x<p$? Taking $x$ to $2^x$ does not give a well-defined function mod $p$. |
Jun 26 |
revised |
Maximum cardinality of a filtered limit of finite sets
added 451 characters in body |