# Is there a notion of torsion for Yang-Mills/gauge connection?

In theories of gravity, the Riemannian/metric connection, is allowed to have torsion, of which the Levi-Civita connection is the particular torsion-free case.

In the gauge theoretic description of QFTs, the gauge connection defines the covariant derivative and the curvature associated to this connection is the field strength.

Is there also a notion of torsion associated to the gauge connection?

• Just a remark: In general, in differential geometry (and in some theories of gravity trying to couple to fermions) torsion is allowed, but in GR, the connection is completely symmetric and torsion is zero. – Dvij Mankad Jun 12 '17 at 9:13
• @Dvij Yes, in Einsteinian GR that's the case. Thanks for pointing it out. Though I meant gravity theories, in general. Edited the question details now. – phydev Jun 12 '17 at 9:28
• Please check this article: https://arxiv.org/abs/1511.07586. Since it is quite lengthy, use the search option of your browser for Yang-Mills. I don't know if it helps but they do talk about it. – PML Jun 12 '17 at 9:45
• Not sure how to define such a thing. Maybe $t-t^T$? A Yang-Mills generator $t$ minus its transpose. – Kosm Jun 12 '17 at 9:52

There isn't.

Torsion represents the failure of "infinitesimal parallelograms" to close.

To see this, consider a point $x$ and and two "infintesimally nearby" points $x_1^\mu=x^\mu+\frac{dx^\mu}{d\tau}d\tau=x^\mu+T^\mu d\tau$ and $x_2^\mu =x^\mu+S^\mu d\sigma$.

Now we parallel transport $T^\mu d\tau$ to the second point $x^\mu+S^\mu d\sigma$:

$$T^\mu(x_2)d\tau=(T^\mu-\Gamma^\mu_{\nu\sigma}S^\nu d\sigma T^\sigma) d\tau,$$ add this to the second point and get $$x_3^\mu= x^\mu +S^\mu d\sigma+T^\mu d\tau-\Gamma^\mu_{\nu\sigma}S^\nu T^\sigma d\sigma d\tau.$$

What if we do the other way around and transport $S^\mu d\sigma$ to $x_1$?

We get $$\tilde{x}_3^\mu=x^\mu+T^\mu d\tau+S^\mu d\sigma-\Gamma^\mu_{\nu\sigma}T^\nu S^\sigma d\sigma d\tau.$$

The difference of these two points is $$\Delta x_3^\mu=(\Gamma^\mu_{\nu\sigma}-\Gamma^\mu_{\sigma\nu})d\sigma d\tau=T^\mu_{\ \nu\sigma}d\sigma d\tau.$$

The reason I say that is to illustrate that the core concept of torsion involves parallel transporting one vector along another, and then doing the same operation in reverse, transporting the "other" vector along the former one. The invariant definition $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$$also implies this.

For a general connection, the direction of parallel transport is given by a vector, but the object to be parallel transported are not vectors but general fiber elements. So the operation of transporting a fiber element along a vector, and then transporting the vector along a fiber element makes no sense, because a fiber element doesn't determine a direction in the manifold in general. Eg. the expression $D_Xs$ cannot be symmetrized or antisymmetrized, since $X$ and $s$ are very different objects.

Curvature doesn't have the same problem, since there two vectors are needed to generate a loop, but only a fiber element gets transported.

No, because the torsion of a connection is defined as the action of this connection on the solder form.

$$T = D \theta$$

The Yang-Mill connection has no solder form associated to it.

• This is not complete; there often several different ways of defining certain objects/quantities, and merely showing that one point of view does not generalize is not enough. – Danu Jun 12 '17 at 10:48