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Let $P\to X$ be a rank $n$ vector bundle with a metric over a smooth manifold $X$, and let $\nabla$ be a connection on $P$ that is compatible with the metric. Fix a point $x\in X$. Then is it true that we can choose an orthonormal basis $e_1,\dots,e_n$ of $P_x$ so that $\nabla=\sum_k \nabla_{e_k}\otimes de_k$ at $x$? I am reading Morgan's book on Seiberg-Witten theory, and in the proof of Proposition 5.1.5, the writer chooses such a basis, but I can't see how this can be done.

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  • $\begingroup$ Your statement makes no sense, as it confuses an orthonormal frame for $P$ with an orthonormal frame for $X$. Moreover, please say what $de_k$ is supposed to mean. See the comments below. $\endgroup$ Jul 25, 2022 at 21:08

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I will assume that the $\{e_i\}$ are a frame for the tangent bundle, as this is the only way in which the expression $\nabla_{e_i}$ could make sense. The connection $1$-form $A$ is defined by $\nabla e_i=\sum e_j\otimes A^j_i$. The connection coefficients $\Gamma^k_{ij}$ are defined by $\nabla_{e_j}e_i=\sum\Gamma^k_{ij}e_k$. We want a basis such that $\nabla e_i=\sum_j\nabla_{e_j}e_i\otimes \theta^j$, denoting by $\theta^j$ the coframe. Using the above definitions, this amounts to $$\nabla e_i=\sum_j(\sum_k\Gamma^k_{ij}e_k)\otimes\theta^j=\sum_ke_k\otimes(\sum_j\Gamma^k_{ij}\theta^j)$$ This means that we want a basis so that the connection $1$-form is related to the connection coefficients via $$A^j_i=\sum_k\Gamma^j_{ki}\theta^k$$ This is the relation between the connection $1$-form and the connection coefficients, for a metric connection (e.g. on https://en.wikipedia.org/wiki/Connection_form#Example:_the_Levi-Civita_connection ). So it should indeed be possible to find such a frame.

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  • $\begingroup$ What does $\theta^j = de_j$ even mean? $\endgroup$ Jul 25, 2022 at 16:49
  • $\begingroup$ Notation for the coframe associated to the frame $e_j$ which is used on wikipedia. $\endgroup$ Jul 25, 2022 at 18:57
  • $\begingroup$ But this is an arbitrary vector bundle, not the tangent bundle. You’re reading things on Wikipedia that do not apply. $\endgroup$ Jul 25, 2022 at 18:58
  • $\begingroup$ Surely, the fact that we have the expression $\nabla_{e_i}$ appearing in the OP means that the $\{e_i\}$ are in fact a frame for the tangent bundle, not for an arbitrary vector bundle? $\endgroup$ Jul 25, 2022 at 19:47
  • $\begingroup$ No, the OP says explicitly that they're an orthonormal frame for $P_x$. I think you need to read questions more carefully. $\endgroup$ Jul 25, 2022 at 20:49

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