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# David Feldman

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 Name David Feldman Member for 2 years Seen Apr 30 at 18:48 Website Location Age
 Jun14 awarded ● Great Question Apr25 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. Apr25 revised An action or two of $SL_2(\Bbb Z)$?added 171 characters in body; added 2 characters in body Apr25 asked An action or two of $SL_2(\Bbb Z)$? Mar19 revised Projective Plane of Order 12Fixed a run on and some spelling. Mar19 revised Projective Plane of Order 12fixed grammar : all other ===> all others Mar18 awarded ● Notable Question Feb28 awarded ● Nice Question Feb28 revised Research level applications of “row rank = column rank”?added 1 characters in body Feb28 revised Research level applications of “row rank = column rank”?added 2 characters in body Feb28 asked Research level applications of “row rank = column rank”? Feb26 awarded ● Nice Question Feb23 comment Not especially famous, long-open problems which higher mathematics beginners can understandYou reference asks for necessary and sufficient conditions that will make the product of Toeplitz operators a Toeplitz operator. Feb23 awarded ● Popular Question Feb21 comment Cantor’s diagonal argument and ZFAsaf, 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. Feb21 revised Cantor’s diagonal argument and ZFedited body Feb21 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. Feb21 comment Cantor’s diagonal argument and ZFHi 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 . Feb19 asked Cantor’s diagonal argument and ZF Feb3 awarded ● Nice Answer Feb3 awarded ● Famous Question Jan17 awarded ● Popular Question