### Exponentiating weak Lie 2-algebras

I have thought a little more about the general problem addressed in the above entry. Maybe it is possible to proceed this way:

While it is hard to deal with weak Lie 2-groups, it is easy to deal with weak (i.e. semistrict) Lie 2-algebras. These are nothing but 2-term ${L}_{\mathrm{\infty}}$ algebras in a different guise and have a simple list of properties. (See for instance lemma 33 of HDA6 or section 2.2 in our ‘From Loop Groups to 2-Groups’).

So here a Lie 2-algebra element (a morphism in the 2-algebra) is a couple

(1)$$(x,\stackrel{\rightharpoonup}{f})$$

with $x\in {V}_{0}$ and $\stackrel{\rightharpoonup}{f}\in {V}_{1}$ elements of two vector spaces, and the bracket on such couples is

(2)$$[({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1}),({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2}),]=({l}_{2}({x}_{1},{x}_{2}),\phantom{\rule{thinmathspace}{0ex}}{l}_{2}({x}_{1},{\stackrel{\rightharpoonup}{f}}_{2})-{l}_{2}({x}_{2},{\stackrel{\rightharpoonup}{f}}_{1})+{l}_{2}(d{\stackrel{\rightharpoonup}{f}}_{1},{\stackrel{\rightharpoonup}{f}}_{2}))$$

where

(3)$${l}_{2}({V}_{0},{V}_{0})={V}_{0}$$

and

(4)$${l}_{2}({V}_{0},{V}_{1})={V}_{1}$$

is the binary operation in the ${L}_{\mathrm{\infty}}$ algebra and

(5)$${V}_{1}\stackrel{d}{\to}{V}_{0}$$

is our 2-term ‘complex’.

Given any such (2-)algebra $\mathcal{L}$ with a bracket we can try to ‘exponentiate’ it to obtain an algebra we might call $\mathrm{exp}(\mathcal{L})$ as follows:

Let the the elements of $\mathrm{exp}\left(\mathcal{L}\right)$ be formal symbols

(6)$$\mathrm{exp}(x,\stackrel{\rightharpoonup}{f})$$

for each $(x,\stackrel{\rightharpoonup}{f})\in \mathcal{L}$ and define the algebra product by means of the Baker-Campbell-Hausdorff formula as

(7)$$\mathrm{exp}({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1})\mathrm{exp}({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2})=\mathrm{exp}(({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1})+({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2})+\frac{1}{2}[({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1}),({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2})]+\frac{1}{12}[({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1})-({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2}),\phantom{\rule{thinmathspace}{0ex}}[({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1}),({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2})]]+\cdots )\phantom{\rule{thinmathspace}{0ex}}.$$

when $\mathcal{L}$ is the 2-term ${L}_{\mathrm{\infty}}$ algebra of a *strict* Lie 2-algebra (which then is nothing but a differential crossed module) then $\mathrm{exp}(\mathcal{L})$ is nothing but the algebra under the horizontal product of a strict 2-group (i.e. a crossed module).

More generally we have a non-vanishing *Jacobiator*

(8)$${l}_{2}(x,{l}_{2}(y,z))+{l}_{2}(z,{l}_{2}(x,y))+{l}_{2}(y,{l}_{2}(z,x))={\mathrm{dl}}_{3}(x,y,z)$$

(for $x,y,z\in {V}_{0}$) and find that the product in $\mathrm{exp}(\mathcal{L})$ is non-associative:

On the one hand we have

(9)$$\left({e}^{\left(x\mathrm{,0}\right)}{e}^{\left(y\mathrm{,0}\right)}\right){e}^{\left(z\mathrm{,0}\right)}=\mathrm{exp}(x+y+z-\frac{1}{12}{\mathrm{dl}}_{3}(x,y,z)+\cdots \phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}0)$$

on the other

(10)$${e}^{\left(x\mathrm{,0}\right)}\left({e}^{\left(y\mathrm{,0}\right)}{e}^{\left(z\mathrm{,0}\right)}\right)=\mathrm{exp}(x+y+z+\frac{1}{12}{\mathrm{dl}}_{3}(x,y,z)+\cdots \phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}0)\phantom{\rule{thinmathspace}{0ex}}.$$

Hence there would be an *associator*

(11)$$({e}^{x}\otimes {e}^{y})\otimes {e}^{z}\stackrel{{\alpha}_{{e}^{x},{e}^{y},{e}^{z}}}{\to}{e}^{x}\otimes ({e}^{y}\otimes {e}^{z})$$

given to lowest order by

(12)$${\alpha}_{{e}^{x},{e}^{y},{e}^{z}}\approx (({e}^{x}{e}^{y}){e}^{z})\mathrm{exp}(0,\frac{1}{6}{l}_{3}(x,y,z))$$

if we (as necessary to get the associative case right) declare the target of $\mathrm{exp}(0,\stackrel{\rightharpoonup}{f})$ to be

(13)$$t\left(\mathrm{exp}(0,\stackrel{\rightharpoonup}{f})\right)={e}^{d\stackrel{\rightharpoonup}{f}}\phantom{\rule{thinmathspace}{0ex}}.$$

I haven’t checked if and how this associator extends to all orders.

One more condition I have checked though is the following: Given the above horizontal product, the exchange law in the weak 2-group to-be uniquely specifies the *vertical product*
of
$\mathrm{exp}({x}_{1},{\stackrel{\rightharpoonup}{f}}_{1})$ with $\mathrm{exp}({x}_{2},{\stackrel{\rightharpoonup}{f}}_{2})$. This, however, has to be associative, since it must be just composition of arrows in a category. By working it out one finds that this implies that

(14)$${l}_{3}(d\stackrel{\rightharpoonup}{f},d\stackrel{\rightharpoonup}{g},\stackrel{\rightharpoonup}{h})=0$$

for all $\stackrel{\rightharpoonup}{f},\stackrel{\rightharpoonup}{g},\stackrel{\rightharpoonup}{h}\in {V}_{1}$.

I am not sure how drastic this restriction cuts down the space of all interesting ${L}_{\mathrm{\infty}}$ algebras, though.

Anyway, if we restrict attention to ${L}_{\mathrm{\infty}}$-algebras that do satisfy this the above might be a way to get a handle on weak 2-groups and hence possibly on weakened nonabelian bundle gerbes.

P.S.

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## Re: Weak NABG

On a related note, does anyone know where the T_A(h) term in the bitorsor connection pullback equation

l^h* a = Ad_h a + p*T_A(h)

in Aschieri-Cantini-Jurco (hep-th/0312154) comes from? Here l^h is left multiplication by h in our group G, a is the Lie(G)-valued 1-form and A is the Lie(Aut(G))-valued 1-form.

I know it’s related to the a_ij in

A_j + a_ij = gA_ig^{-1} + g^{-1}dg

(with the obvious indices on the g’s), but it seems to be pulled out of the air in the above paper. I know we need it to retain left-right symmetry in the bitorsor but the direction ‘see later’ halfway down page 11 isn’t helpful.

David