# Do Carmo Riemannian Geometry, definition 2.6

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 $$\mathbb{R}$$?

• Yes, "function" many times means real-valued functions. Otherwise people call them "maps", like you did yourself Commented Jun 4, 2019 at 22:15
• So f goes from the manifold to R? Can you point out a reference? Commented Jun 4, 2019 at 22:17
• 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
Commented Jun 4, 2019 at 22:32
• Still, a reference would be nice. But I'll read again through the book it might be possible I missed something. Commented Jun 4, 2019 at 22:37

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.