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Martin Sleziak
Martin Sleziak
12:45 PM
It seems that there are users that use the macro \EE.
search comments I found a comment where this macro is not rendered - but I do not see definition anywhere on that page: data.stackexchange.com/math/query/556789/… data.stackexchange.com/math/revision/1066862/1318474/…
No, each individual piece of an $n$-tuple is an element of the field. The idea is that you set up a correspondence between a set of vectors in $V$ and a set of $n$-tuples in $\FF^n$. — Hayden Mar 30 '14 at 22:09
The first query returns one comment - however, in that one the macro is defined:
$\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$I agree that you can directly argue that the map becomes surjective after taking torsion free quotients. But, vaguely speaking, I feel like geometric arguments about moving cycles around are more like trying to give a splitting, which might explain why we can't do it. Taking $M$ orientable doesn't work; as I just wrote over on MO, the diagonal image of $H_3(\RR \PP^3)$ in $H_3(\RR \PP^3 \times \RR \PP^3)$ is not in the image of $\bigoplus H_i \otimes H_j$, and $\RR \PP^3$ is orientable. — David E Speyer Apr 2 '16 at 16:13
I fixed one answer with \RR where the macro was defined in the question: math.stackexchange.com/posts/1662676/revisions
The two comments from the second query seem to be false positives:
$\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$I agree that you can directly argue that the map becomes surjective after taking torsion free quotients. But, vaguely speaking, I feel like geometric arguments about moving cycles around are more like trying to give a splitting, which might explain why we can't do it. Taking $M$ orientable doesn't work; as I just wrote over on MO, the diagonal image of $H_3(\RR \PP^3)$ in $H_3(\RR \PP^3 \times \RR \PP^3)$ is not in the image of $\bigoplus H_i \otimes H_j$, and $\RR \PP^3$ is orientable. — David E Speyer Apr 2 '16 at 16:13
@CameronBuie You edited my answer and now the $\RR$'s are not rendering... — oxeimon Oct 23 '15 at 23:15
And 2) asks: what should $v_i$ be so that $f(v_i)$ has coordinates $(1,0,1,1)$ in the standard base. And if the equation system would be inconsistent here, it doesn't answer the question: does there exist any non-standard base for $\RR^4$ that $f(v_i)$ has these coordinates? — marmistrz Feb 19 '16 at 9:38
I'm asking about a mapping $\RR^4 \rightarrow \RR^3$. ($g(v) = D^T v$) The method for 1) you presented is exactly what I proposed and makes use of the injectiveness. — marmistrz Feb 19 '16 at 9:32
In this case the macro is defined in the question (and I have recently added it to the answer).
\RR is not rendered here - but I do not see definition on the page:
Very nice, thank you. Though you do not need $C_3=[0,1]\times[0,1]$ of course, you should only have $X=\RR^2$ in that place. — Vladimir Sotirov Jun 7 '14 at 16:05
Thank you for the explanation but What is $\RR^n$? — user122358 Mar 19 '17 at 17:49
 

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