I recently heard a differential geometry talk where the speaker constructed a one-parameter family of metrics $g(t)$ on a smooth manifold and said that $g(t)$ is real analytic in the Banach space $BC(T^2(M))$ with respect to the sup norm $\Vert . \Vert_{g(0)}$. I did not understand this at the time, nor can I (with some googling also) make precise sense of this now (I can't guess what $BC$ stands for either). To me, saying something is real analytic means it satisfies a Taylor series expansion. In this situation, I would think that given $x \in M$, $v_1, v_2 \in T_x(M)$, $g(t)$ is real analytic if $$g(t + h)(v_1, v_2) = g(t)(v_1, v_2) + hg(t)(v_1, v_2) + \frac{h^2}{2!}g(t)(v_1, v_2) + ....$$ I don't see why a norm needs to be invoked here. Any insights, please?
1 Answer
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I should just comment but I don't have enough reputation to do that, therefore I write my comment here.
I guess BC stands for "bounded continuous", the sup norm is to indicate what does "bounded" mean. Being analytic w.r.t. $t$ could be considered as its expression (matrices) in the local coordinates, is analytic w.r.t. $t$ exactly what you wrote.
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$\begingroup$ Okay, I see. In case the manifold is not compact, they need bounded continuous to make it a Banach space. Nice observation. $\endgroup$– studentCommented Oct 5, 2015 at 20:47