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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