Musings

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

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

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

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,

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

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

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.