David Feldman
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Registered User
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Jun 14 |
awarded | ● Great Question |
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Apr 25 |
comment |
An action or two of $SL_2(\Bbb Z)$? >$SL(2,Z)$ does not act on the universal cover; the group acting on the universal cover is a certain extension of $SL(2,Z)$ by a free group of infinite rank. Is my mistake only this: $SL_2(Z)$ acts on the universal covering space of $R^2\setminus Z^2 \cup \{(0,0\}$? $SL_2(Z)$ fixes $(0,0)$; I can represent elements of the universal covering space by paths (up to deformation) leaving from $(0,0)$ and terminating wherever. Then $SL_2(Z)$ acts on the paths and homotopies between them. |
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Apr 25 |
revised |
An action or two of $SL_2(\Bbb Z)$? added 171 characters in body; added 2 characters in body |
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Apr 25 |
asked | An action or two of $SL_2(\Bbb Z)$? |
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Mar 19 |
revised |
Projective Plane of Order 12 Fixed a run on and some spelling. |
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Mar 19 |
revised |
Projective Plane of Order 12 fixed grammar : all other ===> all others |
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Mar 18 |
awarded | ● Notable Question |
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Feb 28 |
awarded | ● Nice Question |
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Feb 28 |
revised |
Research level applications of “row rank = column rank”? added 1 characters in body |
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Feb 28 |
revised |
Research level applications of “row rank = column rank”? added 2 characters in body |
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Feb 28 |
asked | Research level applications of “row rank = column rank”? |
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Feb 26 |
awarded | ● Nice Question |
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Feb 23 |
comment |
Not especially famous, long-open problems which higher mathematics beginners can understand You reference asks for necessary and sufficient conditions that will make the product of Toeplitz operators a Toeplitz operator. |
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Feb 23 |
awarded | ● Popular Question |
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Feb 21 |
comment |
Cantor’s diagonal argument and ZF Asaf, what interests me is the ambiguity I find, absent AC, around the word "decreasing." $|A| < |B|$ could mean, minimally, an injection $A\rightarrow B$ (or merely a surjection $B \rightarrow A$), plus the mere absence of any injection $B \rightarrow A$ (or surjection from $A\rightarrow B$). Instead of these "mere absence" conditions, we might demand a uniform supply of witnesses to the failure of any map (which AC would supply if we had it). My comment gives a proof that AC does actually preclude infinite decreasing (in one very strong sense) cardinals. One might seek variations. |
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Feb 21 |
revised |
Cantor’s diagonal argument and ZF edited body |
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Feb 21 |
comment |
Cantor’s diagonal argument and ZF ...this condition. You're right...Cantor's proof gives more, but I wanted to keep the effective burden as light as possible. If in addition $d_n(i)$ depends only on the image of $i$, then an argument along the lines of Ricky Demer's leads to a contradiction (because one gets a nested sub family with a well-ordering). I don't see how to get a contradiction dropping the depends-only-on-image condition, even if one requires witnesses of non-surjectively for all functions, not just injections. So naturally I wondered if any variation on Cantor could meet my side condition. |
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Feb 21 |
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Cantor’s diagonal argument and ZF Hi Joel...Here's what I was originally thinking about (this may become it's own question). ZFC makes cardinals well-ordered. Thus, ($\ast$) given sets $S_1 \supset S_2 \supset \cdots$ there must exist $n$ and a surjection $p:S_{n+1} \rightarrow S_n$. Surely ($\ast$) fails in ZF; I wonder how close one can come. One idea: from an effective guarantee that every injection $h:S{n+1} \rightarrow S_n$ misses an element of $S_n$ implemented by a function $d_n:{\rm Injections}(S_{n+1},S_n)\rightarrow S_n$ such that $d_n(i)\not\in i(S_{n+1})$,try to derive a contradiction. Cantor's proof inspires . |
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Feb 19 |
asked | Cantor’s diagonal argument and ZF |
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Feb 3 |
awarded | ● Nice Answer |
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Feb 3 |
awarded | ● Famous Question |
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Jan 17 |
awarded | ● Popular Question |
