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Bit of trouble understanding the following definition:

Let $M$ be a differentiable manifold. A differentiable function $\alpha : (-\epsilon,+\epsilon)\to M$ is called a differentiable curve in $M$. Suppose $\alpha(0) = p \in M$, and let $\mathcal{D}$ be the set of functions on $M$ that are differentiable at $p$. The tangent vector to the curve $\alpha$ at $t=0$ is a function $\alpha'(0):\mathcal{D}\to \mathbb{R}$ given by $$ \alpha'(0) f = \left( \frac{d(f \circ \alpha)}{dt} \right)_{t=0} $$

The definition continues but it's irrelevant for my question.

Because I don't know the image of the function $f$, but it is assumed to be differentiable shall I assume that $f$ is real valued somehow?

Definition 2.5. defines a mapping $\varphi$ from a manifold to another to be differentiable if its expression is differentiable. The expression is a well defined function from $\mathbb{R}^m$ to $\mathbb{R}^n$. However in the definition above of tangent vector I don't think $f$ is necessarely a function between two manifolds.

The question is how is the definition of $\alpha'(0)$ well defined to be a function from $\mathcal{D}$ to $\mathbb{R}$?

A rural reader
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user8469759
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    Yes, "function" many times means real-valued functions. Otherwise people call them "maps", like you did yourself – Paulo Mourão Jun 04 '19 at 22:15
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    So f goes from the manifold to R? Can you point out a reference? – user8469759 Jun 04 '19 at 22:17
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    It seems pretty clear from context that what Paulo said is true. Just look at how they define $\alpha'(0)$ (both its definition as a mapping and by its action on $f$). – cmk Jun 04 '19 at 22:32
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    Still, a reference would be nice. But I'll read again through the book it might be possible I missed something. – user8469759 Jun 04 '19 at 22:37

1 Answers1

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As an answer to your first question, $f$ is indeed real-valued. Here is a reference on the use of the terms "mapping" and "function" in differential geometry. Even though Do Carmo never says $f$ is real valued explicitly, this is a linguistic convention.

To answer your second question, $$\alpha'(0)f := \displaystyle{\frac{d(f \circ \alpha)}{dt} \Bigg\rvert_{t = 0}}$$ is well defined by virtue of the fact that taking the derivative of a function is itself well defined, as is function composition. If you wanted to break things down into symbols, let $D_0$ denote the differentiation operator with respect to time of real valued functions, evaluated at $t = 0$, and let $C_\alpha$ denote the composition operator, so $C_\alpha(f) = f \circ \alpha$. Then $\alpha'(0)f = D_0 \circ C_\alpha (f)$. Each component of $\alpha'(0)f$ is well defined, and hence $\alpha'(0)f$ is well-defined.

WDR
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