User henry t. horton - MathOverflow

Answer by Henry T. Horton for About the curvature of a connection?

2013-05-22T01:29:58Z

The calculation goes as follows: \begin{align} d_A \circ d_A(\alpha) & = d_A(d\alpha + [A, \alpha]) \\ & = d(d\alpha + [A,\alpha]) + [A, d\alpha + [A,\alpha]] \\ & = d^2 \alpha + d[A, \alpha] + [A, d\alpha] + [A, [A, \alpha]] \\ & = 0 + [dA, \alpha] - [A, d\alpha] + [A, d\alpha] + \tfrac{1}{2}[[A, A], \alpha] \\ & = [dA + \tfrac{1}{2}[A,A], \alpha] \\ & = [d_A A, \alpha]. \end{align} There's a couple basic identities you need to check in the process, but it's nothing difficult. The $\tfrac{1}{2}$ shows up when you use the Jacobi identity, so it seems that perhaps Audin is missing it, but I don't have access to the book currently and can't look at what she does. For matrix Lie groups, $\tfrac{1}{2}[A, A] = A \wedge A$ where we consider $A \wedge A$ as a matrix product, so sometimes you will see $F(A) = dA + A \wedge A$ in the context of matrix Lie groups.
Simply use the definition of the derivative in an affine space (since $\mathcal{A}$ is affine): \begin{align} (T_A F)(\phi) & = \lim_{t \to 0} \frac{1}{t} (F(A + t\phi) - F(A)) \\ & = \lim_{t \to 0} \frac{1}{t}(d_{A + t\phi} (A + t\phi) - d_A A) \\ & = \lim_{t \to 0} \frac{1}{t}(d(A + t\phi) + \tfrac{1}{2}[A, A + t\phi] - dA - \tfrac{1}{2}[A,A]) \\ & = \lim_{t \to 0} \frac{1}{t}(dA + td\phi + \tfrac{1}{2}[A,A] + \tfrac{1}{2}t[A, \phi] - dA - \tfrac{1}{2}[A,A]) \\ & = d\phi + \tfrac{1}{2}[A, \phi] \\ & = d_A \phi. \end{align}

Answer by Henry T. Horton for Helped needed with some characteristic class / number questions

2013-03-29T00:56:04Z

a) How has the Pontryagin class been defined for you? The definition doesn't use an orientation in any way. If $E$ is a real vector bundle (not necessarily oriented) over a manifold $M$, then the Pontryagin classes $p_k(E) \in H^{4k}(M; \Bbb Z)$ can be defined in terms of the Chern classes of the complexification of $E$: $$p_k(E) = (-1)^k c_{2k}(E \otimes \Bbb C).$$ This definition makes no use of any orientation on $E$, so it should be clear that $p_k(E)$ doesn't depend on a choice of orientation.

b) If $E$ is a complex vector bundle and $\bar{E}$ is the conjugate bundle, then we have that $$c_k(\bar{E}) = (-1)^k c_k(E).$$ This is Lemma 14.9 in Milnor and Stasheff's Characteristic Classes.

c) By "diffeomorphic" vector bundles, do you mean isomorphic vector bundles? In that case, yes, characteristic classes are invariants of the isomorphism class of a bundle. One way to see this is by the result you quoted: The classifying maps for two isomorphic bundles are homotopic to each other, and a characteristic class on $E$ is just the pullback of the universal characteristic class on the classifying bundle. Homotopic maps induce the same map on cohomology, so we find that isomorphic bundles have the same characteristic classes.

From b) and c) we can see that the canonical line bundle $\gamma$ and its conjugate bundle $\bar{\gamma}$ are not isomorphic, since $c_1(\gamma) \neq 0$ and $c_1(\gamma) = - c_1(\bar{\gamma})$ ($H^2(\Bbb C P^n; \Bbb Z)$ has no $2$-torsion).

Answer by Henry T. Horton for Wedge Product of Lie Algebra Valued One-Form

2013-03-05T18:26:39Z

There's something about the notation you should know before you get confused when trying to do non-abelian gauge theory. The second term in the field strength should involve a combination of the wedge product of forms and the Lie bracket: the field strength (in the case of an arbitrary gauge group $G$ with Lie algebra $\mathfrak{g}$) should actually be $$F = dA + \tfrac{1}{2}[A \wedge A],$$ where if $\omega$ is a $\mathfrak{g}$-valued $k$-form and $\eta$ is a $\mathfrak{g}$-valued $p$-form, then $$[\omega \wedge \eta](X_1, \dots, X_{k+p}) = \sum_{\sigma \in S_{k+p}} (-1)^{\text{sgn}(\sigma)} [\omega(X_{\sigma(1)}, \dots, X_{\sigma(k)}), \eta(X_{\sigma(k+1)}, \dots, X_{\sigma(k+p)})]$$ for any $k + p$ vector fields $X_1, \dots, X_{k+p}$. In particular, if $A$ is a $\mathfrak{g}$-valued $1$-form, then $$[A \wedge A](X_1, X_2) = [A(X_1), A(X_2)] - [A(X_2), A(X_1)] = 2[A(X_1), A(X_2)].$$ So in components, the field strength is given by $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu],$$ which is the form you'll see most frequently in the physics literature.

When the gauge group $G$ is abelian (e.g. in ${\rm U}(1)$ gauge theory), the Lie bracket on $\mathfrak{g}$ is trivial so that $[A \wedge A] \equiv 0$ and the field strength is just the exterior derivative of the gauge potential: $F = dA$.

Answer by Henry T. Horton for open problems in Seiberg-Witten Theory on 4-Manifolds

2013-02-05T03:09:39Z

One basic problem is determining the relationship between Seiberg-Witten invariants and Donaldson invariants of $4$-manifolds. Witten himself proposed the precise relationship between the two in the original paper Monopoles and 4-Manifolds, but as far as I know the relationship has not been proven in general. Witten's conjecture has been proven in many cases, however. See the answer to this question for a good overview of the current status of this problem.

Answer by Henry T. Horton for Proof of Arnold Conjecture for monotone symplectic manifolds

2013-01-30T00:02:55Z

Note that for the Floer boundary map to be well-defined, we require a Floer regular pair $(H_t, J_t)$ (we want a family of $\omega$-compatible $J_t$ so we have a nice metric with which to take gradients for Floer's equation). So if we're given an $H_t$ to define the Floer homology groups with, we can hope that maybe there is a family $J_t$ of almost complex structures such that $(H_t, J_t)$ is Floer regular. This turns out to be possible -- see for example Exercise 19.22 in Oh's notes (these notes focus on the semipositive case, but monotone symplectic manifolds are semipositive).

Answer by Henry T. Horton for Condition in proof of the Arnold conjecture for monotone manifolds

2013-01-27T18:19:49Z

The normalization $$\int_{S^2} v^\ast \omega \in \Bbb Z \text{ for all smooth } v: S^2 \longrightarrow M$$ is achieved by setting $\omega \mapsto \frac{1}{\tau} \omega$ and looking at the monotonicity condition $$\int_{S^2} v^\ast c_1(M) = \tau \int_{S^2} v^\ast \omega \text{ for all smooth } v: S^2 \longrightarrow M.$$ $c_1(M)$ is an integral class, so after making the substitution $\omega \mapsto \frac{1}{\tau}\omega$ we see that the required integrality condition holds.

Salamon uses this normalization in order to make sure the symplectic action "functional" $$\mathcal{A}_H: \mathcal{L}_0 M \longrightarrow \Bbb R /\Bbb Z,$$ $$\mathcal{A}_H(\gamma) = - \int_{D^2} u^\ast \omega - \int_0^1 H_t(\gamma(t)) ~dt,$$ where $u: D^2 \longrightarrow M$ is such that $u|_{\partial D^2} = \gamma$ , is well-defined up to addition of an integer. We can see that choosing another $u': D^2 \longrightarrow M$ such that $u'|_{\partial D^2} = \gamma$ changes $\mathcal{A}_H(\gamma)$ by an integer amount by "gluing" $u$ and $u'$ along their common boundary to get a map $u \# u': S^2 \longrightarrow M$ and noting that we have $$\int_{S^2} (u \# u')^\ast \omega \in \Bbb Z$$ by our assumption.

The integrality assumption is really just Salamon's way of avoiding the more technical general approach. To get an actual symplectic action functional, you really want to work in the Novikov covering space $\widetilde{\mathcal{L}_0 M}$ of the space of contractible loops in $M$. If you work in this more complicated setting, you get a bona fide symplectic action functional $$\mathcal{A}_H: \widetilde{\mathcal{L}_0 M} \longrightarrow \Bbb R,$$ $$\mathcal{A}_H([\gamma, u]) = - \int_{D^2} u^\ast \omega - \int_0^1 H_t(\gamma(t)) ~dt.$$ You don't need the normalization once you're working with the symplectic action functional on the Novikov covering space.

Salamon briefly mentions the Novikov covering space on page 41 of his notes. A much more detailed reference for this is chapter 18 of Oh's notes.

Answer by Henry T. Horton for decompose a connection

2013-01-08T15:16:15Z

Check out Kobayashi and Nomizu's Foundations of Differential Geometry, Volume 1. On page 159, when proving the existence of the Levi-Civita connection (Theorem IV.2.2), they pick an arbitrary metric connection and add the contorsion tensor to it and show that it is a metric connection with vanishing torsion. Hence any metric connection can be written as the difference of the Levi-Civita connection and its contorsion tensor.

Another reference is Section 7.2.6 of Nakahara's Geometry, Topology, and Physics. See equations (7.30)-(7.35) for Nakahara's derivation.

Answer by Henry T. Horton for What is geometrically the Pontryagin class?

2012-12-22T19:42:13Z

Some fractional Pontrjagin classes are obstructions to higher analogues of orientations/spin structures.

For example, a spin vector bundle $E \longrightarrow X$ admits a string structure if $\frac{1}{2}p_1(E) = 0$. In other words, a spin structure on $E$ determines a class $\lambda = \frac{1}{2} p_1(E) \in H^4(X; \mathbb{Z})$ such that $2\lambda = p_1(E)$, and this fractional first Pontrjagin class $\lambda$ is the obstruction to the existence of a string structure on $E$.

Similarly, if we go to the next nontrivial step on the Whitehead tower, we can try to define a so-called fivebrane structure on a string vector bundle $E \longrightarrow X$. In this case, the obstruction to the string vector bundle $E \longrightarrow X$ admitting a fivebrane structure is the fractional second Pontrjagin class $\frac{1}{6}p_2(E)$.

Answer by Henry T. Horton for Is the wedge sum of two cones over the hawaiian earring contractible?

2012-12-21T19:51:44Z

See the proof of Theorem 2.6 of

J.W. Cannon, G.R. Conner, The combinatorial structure of the Hawaiian Earring group, Topology Appl. 106 (3) (2000) 225–271.

A copy of this paper is available on the second author's webpage here.