1
$\begingroup$

I'm a little confused by the following computation from Baez and Muniain's Gauge Fields, Knots, & Gravity.

Let $M = [0,1]\times S$ where $S$ is any compact oriented manifold. Let $E \overset{\pi}{\to} M$ be a trivial vector bundle over $M$ with connection $A$ in ``temporal gauge", i.e if $t$ is the coordinate corresponding to $[0,1] \subseteq M$, then $A(\frac{\partial}{\partial t}) = 0$. We resolve the exterior derivative as $d = dt\wedge\partial_t + d_S$.

Why is $$2\int_M\text{tr}(dt\wedge\partial_tA\wedge d_sA + dt\wedge\partial_t A\wedge A \wedge A) = \int_M\text{tr}\Big(dt\wedge\partial_t(A\wedge d_sA + \frac{2}{3}A\wedge A\wedge A)\Big)~?$$

It is only the first part of the RHS that I don't understand. i.e. I belive the the correspondence $$2\int_M\text{tr}(dt\wedge\partial_tA\wedge d_sA) = \int_M\text{tr}\big(dt\wedge\partial_t(A\wedge d_sA)\big)$$ is off by a factor of two on the RHS.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Let me use some shorthand notations (which are very popular among physicists) $$d_t = dt \wedge \partial_t$$ Secondly I will not write the wedges. All the multiplications are to be understood as (noncommutative) wedge products.

In addition set the total exterior derivative over $I\otimes M$ $$d = d_t + d_s$$ Since $d$, $d_t$, $d_s$, are exterior derivatives over manifolds, we have: $$ d^2 = d_t ^2 = d_s^2 = 0$$ Therefore $$ d^2 = d_t^2 + d_t d_s + d_s d_t + d_s^2 = 0$$ Implies: $$ d_s d_t + d_t d_s = 0$$ Also we need to remember the general cyclic trace identity ($B$ and $C$ are Lie algebra valued forms). $$\mathrm{tr} (B \wedge C) = (-1)^{\mathrm{rank}(B)\mathrm{rank}(C)} \mathrm{tr} (C\wedge B)$$

The first term $$ \begin{align*} \mathrm{tr} (d_t A d_s A) &= \mathrm{tr} d_t ( A d_s A)- \mathrm{tr} (A d_t d_s A)\\ &= \mathrm{tr} (d_t ( A d_s A))+ \mathrm{tr} (A d_s d_t A) \\ &= \mathrm{tr} d_t ( A d_s A)+\mathrm{tr}( d_s (A d_t A)) - \mathrm{tr} (d_s A d_t A)\\ &= \mathrm{tr} d_t ( A d_s A)+d_s\mathrm{tr}(A d_t A) - \mathrm{tr} (d_t A d_s A) \end{align*} $$

Therefore:

$$2 \mathrm{tr} (d_t A d_s A) = \mathrm{tr} d_t ( A d_s A)+d_s\mathrm{tr}(A d_t A)$$

The second term is an exact form (total derivative) over a compact manifold $M$ therefore it is identically zero. Therefore:

$$2 \mathrm{tr} (d_t A d_s A) = \mathrm{tr} d_t ( A d_s A)$$

The second term $$\mathrm{tr} (d_t ( A^3) )= \mathrm{tr} ((d_t A)A^2 )- \mathrm{tr} (A(d_t A)A )+ \mathrm{tr} (A^2 (d_t A)) $$ Using the cyclic trace rule mentioned above, the second term has an additional $-1$ contribution, thus we have: $$\mathrm{tr} (d_t ( A^3) )= 3 \mathrm{tr} ((d_t A)A^2$$

Remark: It is possible to use the trace rule because both $d_tA$ and $A^2 = A \wedge A$ are Lie algebra valued forms.

Improve this answer
$\endgroup$
2
  • $\begingroup$ This is great, thanks! The only thing is I think we should have $$\text{tr}(d_tAd_sA) = \text{tr}d_t(Ad_sA) + \text{tr}(Ad_td_sA)$$ since $$d_t(Ad_sA) = d_tAd_sA + (-1)^1 Ad_td_sA$$ then the second term equals $$\text{tr}(Ad_td_sA) = -\text{tr}(Ad_sd_tA) = -\text{tr}d_s(Ad_tA) - \text{tr}(d_sAd_tA)$$ everything else proceeds as given $\endgroup$
    – user140223
    Commented Aug 2, 2017 at 18:01
  • 1
    $\begingroup$ Indeed, you are right. Thank you for the correction. By the way, a good place to further exercise similar manipulation of Lie algebra valued forms, and further techniques, is in the theory of anomalies. Please see the following article projecteuclid.org/download/pdf_1/euclid.cmp/1104114308 by Mañes, Stora and Zumino $\endgroup$
    – David Bar Moshe
    Commented Aug 3, 2017 at 6:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.