Theo Johnson-Freyd
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 6h awarded Nice Question 6h comment Does $E_8$ know $Spin(7)$? @VítTuček I don't know how canonical the passage $Spin(7) \leadsto Ham(8,4)$ --- I agree it looks like it needs a basis, but there might be a canonical way to do it (akin to choosing the unique-up-to-conjugation Cartan). The part I'm hoping to understand is $Ham(8,4) \leadsto Spin(7)$ or $E_8 \leadsto Spin(7)$ (I assume that $E_8 \leadsto Ham(8,4)$ isn't too bad, as some sort of reduction mod 2). Actually, I'd like to understand this machine well enough that I can feed it the Golay code (Leech lattice). 7h revised Does $E_8$ know $Spin(7)$? added 50 characters in body 7h comment Does $E_8$ know $Spin(7)$? @VítTuček Oops, yes. I will fix. 11h asked Does $E_8$ know $Spin(7)$? Apr 26 revised When are Morita classes represented by certain structured algebra objects? added 545 characters in body Apr 26 answered When are Morita classes represented by certain structured algebra objects? Apr 24 comment Is the antipode anti-bracketed? In Sweedler notation, $S_\hbar(x) = S(x) - \hbar S(x_{(1)}) \{x_{(2)},S(x_{(3)})\}$. This should do the trick. Apr 24 comment Is the antipode anti-bracketed? Ok, so then I want to solve $u\circ \epsilon = \mathrm{id} \star S + \hbar \mathrm{id} \star' S + \hbar \mathrm{id} \star S'$, i.e. $\mathrm{id}\star S' = -\mathrm{id} \star' S$. But $S \star \mathrm{id} = u \circ \epsilon$, and $u \circ \epsilon$ is the identity for $\star$, so $S' = (u\circ \epsilon) \star S' = (S \star \mathrm{id}) \star S' = S \star (\mathrm{id} \star S')$ by associativity of $\star$, and we wanted to solve $\mathrm{id} \star S' = - \mathrm{id} \star' S$. So we get a formula for $S'$, namely $S' = - S \star (\mathrm{id} \star' S)$. Apr 24 comment Is the antipode anti-bracketed? Here's the fast thing to say. Recall the convolution $\star$. Define the deformed convolution $f\star_\hbar g = \cdot_\hbar \circ (f\otimes g) \circ \Delta$ for $f,g$ linear maps, so that $f\star_\hbar g : x \mapsto f(x_{(1)}) g(x_{(2)}) + \hbar \{ f(x_{(1)}),g(x_{(2)})\}$. Then I can write $\star_\hbar = \star + \hbar \star'$, where $f\star' g = (\{,\}) \circ (f \otimes g) \circ \Delta$. Now I want to solve $\mathrm{id} \star_\hbar S_\hbar = u \circ \epsilon$, where $u : \mathbb C[\hbar]/\hbar^2 \to A[\hbar]/\hbar^2$ is inclusion of the unit. Apr 24 comment Is the antipode anti-bracketed? ... $\{x_{(1)},S(x_{(2)})\} + x_{(1)} S'(x_{(2)}) = 0$. This is supposed to define the map $S'$. ($S'$ was just a name for the second component, but it plays the role of $\partial S_\hbar / \partial \hbar$, so in that sense the $'$ is deserved.) Apr 24 comment Is the antipode anti-bracketed? Let's see. The rule for the antipode is usually $x_{(1)} \cdot S(x_{(2)}) = x$, where in Sweedler notation $\Delta(x) = x_{(1)} \otimes x_{(2)}$, and I don't write the $\sum$ sign (but it is a nontrivial sum). So now let $S_\hbar$ denote the sought-after deformed antipode; I want to solve $x_{(1)} \cdot_\hbar S_\hbar(x_{(2)}) = x$, and it suffices to assume $x\in A \subseteq A[\hbar]/\hbar^2$ by $\mathbb C[\hbar]/\hbar^2$-linearity. By sending $\hbar\to 0$, you see that $S_\hbar(x) = S(x) + S'(x)\hbar$ for $S$ the undeformed antipode. Then the nontrivial part of the equation is ... Apr 24 comment Is the antipode anti-bracketed? The unit is $1$. The counit if $\epsilon$. These do not deform. (Note that I am thinking of deformed algebra $A[\hbar]/\hbar^2$ as a Hopf algebra over the dual numbers $\mathbb C[\hbar]/\hbar^2$, so $\epsilon(a+b\hbar) = \epsilon(a) + \epsilon(b)\hbar$. The point about "it's an open condition" is that you can solve for $S$. Apr 23 answered Is the antipode anti-bracketed? Apr 20 accepted What is the Heegaard Floer Homology of a connect sum of $S^2 \times S^1$s? Apr 20 asked What is the Heegaard Floer Homology of a connect sum of $S^2 \times S^1$s? Apr 19 awarded Favorite Question Apr 16 comment The Quantum Group ${\cal O}_q(SL(n))$, for $q>1$ Is this just $q=p^{-1}$? Apr 14 comment When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$? @benmckay If this question were asked now, as a sociological fact I would agree with you. But please note the date of the question. Apr 13 awarded Notable Question