Timeline for What is growth of ass. algebra with 3 generators and relation a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 ?
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| Mar 8, 2012 at 17:38 | vote | accept | Alexander Chervov | ||
| Nov 22, 2011 at 13:45 | comment | added | Alexander Chervov | @David Sorry I cannot understand. Consider u1 = a3 - "lexigraphically biggest", take v2=a3 then u1(a1a2a3 )= a3a1a2a3 = (a3a1a2)a3 = (a3a1a2)v2 . So the u1(a1a2a3 ) is contained in $\Delta$v1 . This contradicts the claim "monomial only occurs in the term and no others". Am I wrong ? | |
| Nov 21, 2011 at 9:56 | comment | added | Alexander Chervov | @David Thank You very much for Yours answer and comments. I still need some time to get what you wrote, but any way this is very very interesting and helpful for me ! | |
| Nov 20, 2011 at 16:54 | comment | added | David E Speyer | Certainly it can. Take each of those $123$'s and replace them by $213+132-231-312+321$, then expand out to get $25$ terms. | |
| Nov 20, 2011 at 16:41 | comment | added | M T | I don't think you can assume that. Something like $123\Delta 123$ is a relation, but it can't be written as a combination of $u_i\Delta v_i$s if the $u$s and $v$s are 123-free. | |
| Nov 20, 2011 at 16:37 | comment | added | David E Speyer | Uggh, you're right, which means I need to go back and give the step that I thought I could skip. OK, so insert the following into my first comment -- since it is obvious that the standard monomials span, we can assume that the $u_i$ and $v_i$ are standard. Does that fix it? | |
| Nov 20, 2011 at 16:35 | comment | added | M T | Say $u_1=123$ (I'll drop the $a$s). Then if $v_1=123123$ we have in $u_1\Delta v_1$ the monomial $u_1 123 v_1 = 123\cdot 123 \cdot 123123$, and you say this can't appear elsewhere. But if $u_2=123123$, bigger than $u_1$ in the lex order, and $v_2=123$, then a term in $u_2\Delta v_2$ is $123123 \cdot 123 \cdot 123$. Or am I misunderstanding? | |
| Nov 20, 2011 at 16:27 | comment | added | David E Speyer | This is the "standard" noncommutative Groebner basis argument. As always, I am terrible for references. If you are good at references, please post one! | |
| Nov 20, 2011 at 16:26 | comment | added | David E Speyer | For concreteness, I'll take my term order to be lexicographic. Let $u_1$ be the lexicographically first $u_i$ and, if that monomial occurs as $u_i$ more than once, choose the one so that $v_i$ is lexicographically earliest. Then the monomial $u_1 (a_1 a_2 a_3) v_1$ only occurs in the term $u_1 \Delta v_1$ and no others, so it can't cancel out. So the right hand of the sum contains the nonstandard term $u_1 (a_1 a_2 a_3) v_1$, a contradiction. | |
| Nov 20, 2011 at 16:25 | comment | added | David E Speyer | Let $\Delta$ be the defining cubic. Suppose for contradiction that the standard monomials are not linearly independent in degree $n$. So we have $\sum b_i w_i = \sum c_i u_i \Delta v_i$, where $w_i$ are standard monomials, $b_i$ and $c_i$ are nonzero constants and the $u_i$ and $v_i$ are monomials. Continued... | |
| Nov 20, 2011 at 16:02 | comment | added | M T | How do you show that the images of these monomials are linearly independent in this algebra? | |
| Nov 20, 2011 at 15:52 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Nov 20, 2011 at 15:41 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Nov 20, 2011 at 15:32 | history | answered | David E Speyer | CC BY-SA 3.0 |