Timeline for Clifford algebra non-zero
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 11, 2011 at 11:17 | comment | added | DamienC | I made have misubderstood something, but I don't think it is wrong. One does not need $R$ to sit inside the alternating tensors to get that $R\subset V\otimes V\subset A\otimes A\to A$ is zero (it is zero just because $A=T(V)/R$, by definition). | |
| Jul 11, 2011 at 11:17 | comment | added | darij grinberg | Thanks a lot for the explanation - now it is finally clear to me! | |
| Jul 11, 2011 at 10:17 | comment | added | DamienC | Sorry for replying so late. The point is that one has to observe that the following composed map is zero: $R\subset V\otimes V\subset A\otimes A\to A$ (the last map is the prouct in $A$). This implies in particular that when restricted to $\tilde{K}^i(A)$, the differential of $B^\cdot(A)$ has only two non-tricial terms: if $a\otimes v_1\otimes\cdots\otimes v_i\otimes b$ (sum implicitely assumed) lies in $\tilde{K}^i(A)$, then its differential is given by $$ a\cdot v_1\otimes v_2\cdots\otimes v_i\otimes b+(-1)^{i} a\otimes v_1\otimes\cdots\otimes v_{i-1}\otimes v_i\cdot b $$ Do you agree? | |
| Jul 6, 2011 at 15:17 | comment | added | darij grinberg | Sorry for necromancing this up now, but my exams are over and I finally have some real time. It seems that the point where I stop understanding Braverman/Gaitsgory is earlier: in §3.6, why is $\widetilde{K}^{\cdot}\left(A\right)$ a subcomplex of $B^{\cdot}\left(A\right)$ ? As far as I understand, the differential of $B^{\cdot}\left(A\right)$ involves multiplying together adjacent tensorands, which leads us out of the submodule $\widetilde{K}^i\left(A\right)$... What am I doing wrong? | |
| Jun 21, 2011 at 17:55 | comment | added | DamienC | $\alpha\otimes id-id\otimes\alpha$ defines a map from $K^3$ to $T(V)$, that induces (by composing with the projection $T(V)\to A$) a map $K^3\to A$, that extend to an $A$-bimodule map $A\otimes K^3\otimes A=\tilde{K}^3\to A$. | |
| Jun 21, 2011 at 17:13 | comment | added | darij grinberg | How is $\alpha\otimes id-id\otimes\alpha$ viewed as an $A$-bimodule map to $A$? I'd expect it to go to $A\otimes A$ (if not to something larger), or do you compose it with multiplication? Also, how exactly is $d$ defined in $d\left(\widetilde{\alpha}\right)$ ? Is it the usual coboundary in the sense of $d\left(\widetilde{\alpha}\right)=\widetilde{\alpha}\circ\left(\mu\otimes id\otimes id\otimes id - id\otimes\mu\otimes id\otimes id + id\otimes id\otimes\mu\otimes id - id\otimes id\otimes id\otimes\mu\right)$ (where $\mu$ is the multiplication map in $A$)? | |
| Jun 21, 2011 at 15:18 | comment | added | DamienC | $\alpha:K^2=R\to V\subset A$ is viewed as an $A$-bimodule map $\tilde{\alpha}:\tilde{K}^2=A\otimes R\otimes A\to A$. In the very same way you view $\alpha\otimes id-id\otimes\alpha$ as an $A$-bimodule map $\tilde{\gamma}:\tilde{K}^3\to A$. It happens (computation) that $d(\tilde{\alpha})=\tilde{\gamma}$. Now $\alpha\otimes id-id\otimes\alpha$ takes values in $R$ iff $\gamma=0$. | |
| Jun 21, 2011 at 14:54 | comment | added | darij grinberg | Yes, my result is general... but if Braverman-Gaitsgory's ones generalize to relative homology, they will probably achieve the same generality as mine and even more. I am still trying to understand their argument, but I am hitting brick walls. Do you happen to know why (i) is equivalent to $d\alpha = 0$ in §4.3? | |
| Jun 21, 2011 at 14:48 | history | answered | DamienC | CC BY-SA 3.0 |