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Working through a proof that any connected orientable manifold has two orientations (From Tu's manifolds), I'm having a difficulty with a small nuance in the proof. Consider a manifold $ M $ with two orientations $ \mu $ and $ \nu $, and some arbitrary point $ p \in M $. $ \mu_p $ and $ \nu_p $ are the same or opposite orientations of $ T_p M $. The strategy of the proof is to show that the function $ f:M \rightarrow \{1,-1\} $ (where $ f(p) = 1 $ if $ \mu_p = \nu_p $ and $ f(p) = -1 $ if $ \mu_p = - \nu_p $) is locally constant, and therefore globally constant since $ M $ is connected.

Okay, all well and good. Now, the orientations are continuous, so for any point there is a connected open neighborhood of $ p $, $ U $, where $ \mu = [(X_1,...,X_n)] $ and $ \nu = [(Y_1,...,Y_n)] $ are equivalence classes of continuous frames on $ U $. Of course, there is a transition matrix between these frames, $ A:U \rightarrow GL(n,\mathbb{R}) $ with continuous entries, meaning the nonsingular $ \text{det} A:U \rightarrow \mathbb{R}^\times $ is continuous.

Here's the kicker, though: Tu says that because of the Intermediate Value Theorem, $ (\text{det} A)(p) $ is always positive or always negative on $ U $, being unable to change signs without smoothly crossing over zero---thus showing that $ f $ is constant on U. However, I have no idea what the intermediate value theorem means when not talking about functions from some closed interval of $ \mathbb{R} $ to $ \mathbb{R} $. Naturally, I'd say $ \text{det} A $ being continuous means that $ \text{det} A \circ \phi^{-1}:\phi(U) \rightarrow \mathbb{R} $ is continuous for a chart $ (U,\phi) $. But, how is the intermediate value theorem extended to a function $ \mathbb{R}^n \rightarrow \mathbb{R} $? My intuition is that, like in one dimension, the connectedness of $ U $ is key to an analogous theorem, but can a generalization of the IVT be made without introducing notions of curves along $ U $ and other differential geometry machinery into the mix? Can I apply the 1-D IVT to all possible curves between two points in $ U $ or something? I get the intuitive sense of what he means by it, but how can I put it on more mathematical footing?

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2 Answers 2

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The determinant is a continuous function from a connected open set to $\mathbb{R} -\{0\}$, so its image is a connected set.

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  • $\begingroup$ And $ \mathbb{R} - \{0 \} $ has two connected components, so the image can only be one of those, I see. Thanks for humoring me! $\endgroup$
    – Jerome
    Commented Dec 20, 2020 at 23:17
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Usually the IVT is understood as a theorem of elementary calculus which deals with functions $f : [a,b] \to \mathbb R$. However, there is a generalized version.

Note that if $f : X \to Y$ is a continuous map and $X$ is connected, then $f(X)$ is connected. If $Y = \mathbb R$, then the connected subsets of it are precisely the intervals. Therefore

Let $f : X \to \mathbb R$ be a continuous map and $X$ be connected. If $y_1 = f(x_1) < y_2 = f(x_2) $, then for each $y \in (y_1,y_2)$ there exists $x \in X$ such that $f(x) = y$.

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  • $\begingroup$ I see. I suppose it is enough that $ U $ is connected, so $ \text{det} A $ maps to one of the connected halves of the separation $ \mathbb{R}^{\times} = (- \infty, 0) \cup (0, \infty) $, but it is possible to specialize the result to something like an IVT. I suppose I just got hung up on the specific usage of IVT, which wasn't really necessary to even state in the proof. Thanks much! $\endgroup$
    – Jerome
    Commented Dec 20, 2020 at 23:15

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