## May 12, 2008

### Superconnections for Dummies

For inexplicable reasons, I got involved in a discussion with Urs Schreiber about Quillen superconnections. Urs was enamoured of the idea that Quillen superconnections might be relevant to Garrett Lisi’s “Theory of Everything.” Dubious applications, aside, Urs wanted to construct Quillen superconnections from $\mathbb{Z}_2$ graded Lie algebras, instead of Lie superalgebras. This just plain doesn’t work. So while there is, perhaps, some kind of superconnection of the sort he’s after, it’s certainly not the Quillen superconnection.

So here’s an elementary, “For Dummies,” guide to superconnections.

Let $\mathfrak{g}=\mathfrak{g}^0 + \mathfrak{g}^1$ be a Lie superalgebra. The generators satisfy (anti)commutation relations

(1)\begin{aligned} [T_a, T_b] &= \tensor{f}{_a_b_^c}T_c\\ [T_a, \tau_\alpha] &= \tensor{c}{_a_\alpha_^\beta}\tau_\beta\\ \{\tau_\alpha, \tau_\beta\} &= \tensor{d}{_\alpha_\beta_^a}T_a \end{aligned}

which, in turn, satisfy certain associativity conditions called the super-Jacobi identity.

$\mathfrak{g}^0$ (spanned by the $T_a$) is an ordinary Lie algebra. Let $G_0$ be the corresponding Lie group, and $P\to X$ a $G_0$ principal bundle. $\mathfrak{g}$ furnishes a representation of $G_0$. Let $\mathbb{G}= \mathbb{G}^0 \oplus \mathbb{G}^1$ be the corresponding $\mathbb{Z}_2$ graded vector bundle associated to $P$. The main applications, to K-theory (or KO-theory) will involve $\mathfrak{g}= u(n|m)$ (or $so(n|m)$), but we can write the formulæ below for any Lie superalgebra.

Now let $V = V^0 \oplus V^1$ be any representation of $\mathfrak{g}$ (a $\mathbb{Z}_2$ graded vector space). Let $E=E^0\oplus E^1$ be the corresponding $\mathbb{Z}_2$ graded vector bundle associated to $P$. A Quillen superconnection on $E$ is given by the formula

(2)$\mathbb{D} = d + dx^\mu A^a_\mu(x) T_a + \phi^\alpha(x) \tau_\alpha$

where $A$ is an ordinary $G_0$ connection on $E$, and $\phi\in \Gamma(X, \mathbb{G}^1)$. Note that, though $\phi$ is an “odd” section of $\mathbb{G}$, both $A_\mu^a(x)$ and $\phi^\alpha(x)$ are bosonic fields.

The curvature of $\mathbb{D}$ is

(3)\begin{aligned} \mathbb{F} =\mathbb{D}^2= &\tfrac{1}{2} dx^\mu \wedge dx^\nu (\partial_\mu A_\nu-\partial_\nu A_\mu + \tensor{f}{_b_c_^a} A^b_\mu A^c_\nu)T_a\\ & +\, dx^\mu (\partial_\mu \phi^\alpha + \tensor{c}{_a_\beta_^\alpha}A^a_\mu \phi^\beta)\tau_\alpha\\ & +\, \tfrac{1}{2} \tensor{d}{_\alpha_\beta_^a} \phi^\alpha\phi^\beta T_a \end{aligned}

Note that the 0-form term (the last line of (3)) only makes sense because (1) is a Lie superalgebra, and hence $\tensor{d}{_\alpha_\beta_^a}$ is symmetric on $\alpha,\beta$.

In the applications to K-theory, $E^0$ and $E^1$ are the bundles living on the $D9$ and $\overline{D9}$ branes, and $\phi$ are the tachyons that condense between them.

Urs wanted to replace the Lie superalgebra (1) by a $\mathbb{Z}_2$-graded Lie algebra,

(4)\begin{aligned} [T_a, T_b] &= \tensor{f}{_a_b_^c}T_c\\ [T_a, t_\alpha] &= \tensor{c}{_a_\alpha_^\beta}t_\beta\\ [t_\alpha, t_\beta] &= \tensor{\tilde{d}}{_\alpha_\beta_^a}T_a \end{aligned}

The reason he wants to do this is that $e_8$ is a Lie algebra, not a Lie superalgebra, and it admits various $\mathbb{Z}_2$ gradings.

However, because $\tensor{\tilde{d}}{_\alpha_\beta_^a}$ is now antisymmetric on $\alpha,\beta$, we can’t use this to construct a Quillen superconnection (there’s a longer-winded explanation in my discussion with Urs).

But there is another kind of superconnection, let’s call it a “Schreiber superconnection”, which we can build, based on a $\mathbb{Z}_2$-graded Lie algebra. Let

(5)$\mathbb{D} = d + dx^\mu A^a_\mu(x) T_a + \psi^\alpha(x) t_\alpha$

where now $\psi$ is a fermionic field. Its curvature

(6)\begin{aligned} \mathbb{F} =\mathbb{D}^2=&\tfrac{1}{2} dx^\mu\wedge dx^\nu (\partial_\mu A_\nu-\partial_\nu A_\mu + \tensor{f}{_b_c_^a} A^b_\mu A^c_\nu)T_a\\ & +\, dx^\mu (\partial_\mu \psi^\alpha + \tensor{c}{_a_\beta_^\alpha}A^a_\mu \psi^\beta)t_\alpha\\ & +\, \tfrac{1}{2} \tensor{\tilde{d}}{_\alpha_\beta_^a} \psi^\alpha\psi^\beta T_a \end{aligned}

makes sense because, though $\tensor{\tilde{d}}{_\alpha_\beta_^a}$ is antisymmetric on $\alpha,\beta$, the $\psi$ are fermionic.

In Quillen’s case, “$dx^\mu$” and “$\tau_\alpha$” are anti-commuting (“fermionic”), so that $\mathbb{D}$ is overall fermionic, and $\mathbb{F}$ is bosonic. For the “Schreiber superconnection,” “$dx^\mu$” and “$\psi^\alpha(x)$” are anticommuting, with the same net parities for $\mathbb{D}$ and $\mathbb{F}$ as before.

Now, when all the dust settles, the Schreiber superconnection is equally useless for Lisi’s purposes as the Quillen superconnection (though for different reasons). But, at least we’re not taking the name of Quillen in vain.

Posted by distler at May 12, 2008 12:01 AM

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### Re: Superconnections for Dummies

Hi Jacques,

A small typo: in the LHS of the second line of eq (1), $\tau_\beta$ should be $\tau_\alpha$.

Posted by: Lord Sidious on May 12, 2008 12:39 AM | Permalink | Reply to this

### Typos

Not the only such typo. Hopefully they’re all fixed now.

Thanks!

Posted by: Jacques Distler on May 12, 2008 4:44 AM | Permalink | PGP Sig | Reply to this

### Re: Superconnections for Dummies

Hi Jacques,

in my post and in the comments I suggested to interpret the expression

$d + dx^\mu A^a_\mu(x) \rho(T_a) + \phi^a(x) \rho(t^a)$

with (!)

$\rho : g \to end(V)$

a $\mathbb{Z}_2$-graded representation as a Quillen superconnection proper, simply observing that any sum of an ordinary connection on a $\mathbb{Z}_2$-graded vector bundle with a 0-form with values in odd endomorphisms can be read as a Quillen superconnection (Dumitrescu takes that as the definition). In this context your $g$ equals $end(V)$, but not my $g$.

The main point here was that it is not as weird to consider such mixed sums as some discussion around the blogs and elsewhere have implied.

What you call a Schreiber superconnection is more like what Garrett Lisi seemed to have in mind himself, who addresses his “$\psi$” as Grassmannian. But it’s hard to tell, because the difference in the Quillen and the Schreiber superconnection is just in the quadratic term, which he doesn’t write.

But in any case, sorry if indeed I added to the confusion where I was hoping to reduce it. Seems to be one of the things which are easier to clarify in personal discussion than in (inter-)blog discussion.

Finally, I should add that I am not after justifying any problematic speculations here, but that I do find it interesting to ponder (if only for my own humble education) in general what connections “with values in $e_8$” are like, if we allow the “fermionic contribution” to the connection (under your favorite splitting) to come from 0-forms (either way).

Posted by: Urs Schreiber on May 12, 2008 2:52 PM | Permalink | Reply to this

### Re: Superconnections for Dummies

In this context your $g$ equals $end(V)$, but not my $g$.

No, my $\mathfrak{g}$ is a subalgebra of $end(V)$. My point is that you really do need the algebra structure, not just $\mathfrak{g}$ as a $\mathbb{Z}_2$-graded vector space.

For instance, in constructing the curvature, $\mathbb{F}$, from the superconnection, $\mathbb{D}$, we used the algebra structure. And (as I hope my exposition makes clear) a Lie superalgebra is the structure that we need.

What you call a Schreiber superconnection is more like what Garrett Lisi seemed to have in mind…

I agree.

Which is why we should be pleased by the result, because whereas the Quillen superconnection is naturally associated with Lie superalgebras, the Schreiber superconnection (with its fermion field) is naturally associated to $\mathbb{Z}_2$-graded Lie algebras.

… the quadratic term, which he doesn’t write.

He claims (incorrectly, as usual) that it vanishes by Fermi statistics.

I do find it interesting to ponder … if we allow the “fermionic contribution” to the connection … to come from 0-forms.

There’s an important point here, so let me reiterate it. Lisi is not interested in just any old $\mathbb{Z}_2$-grading of $e_8$. He wants a very particular one. Namely, choose some embedding of $SL(2,\mathbb{C})\hookrightarrow E_8$. This defines an action of $SL(2,\mathbb{C})$ on $e_8$. He wants the $\mathbb{Z}_2$-grading that comes from the action of the ($\mathbb{Z}_2$) center of $SL(2,\mathbb{C})$ on $e_8$.

Then it’s automatic that the $\mathbb{Z}_2$-odd generators transform as spinors of $SL(2,\mathbb{C})$.

Posted by: Jacques Distler on May 12, 2008 3:32 PM | Permalink | PGP Sig | Reply to this
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