# Understanding dual tangent space using the definition $T_p \to \mathbb{R}$?

I'm working through Carroll's book, in which he defines a dual space to a tangent space $T_p$ as a mapping $\omega: T_p \to \mathbb{R}$. Now, if we define some chart on the manifold, we can expand an arbitrary element of the tangent space $T_p$ as

$$\frac{d}{d\lambda} = v^\mu \hat{e}_{(\mu)} \in T_p.$$

Clearly, this is not a scalar, but we need to find some object $\omega$ such that its action makes it a scalar. The first thing I can think of that fulfills this role is the right action of $\omega = f$, a scalar function itself. However, Sean Carroll precisely states that $f$ should not be thought of as a one form. Instead, we define the left action of the gradient of $f$ as the one form. From (1.52) we may write

$$\text{d}f \frac{d}{d\lambda} = (\partial_\sigma f)\hat{\theta}^{(\sigma)}v^\mu\hat{e}_{(\mu)}.$$

Now, I see how this proof could be completed and yield $\frac{df}{d\lambda}$--what we would attain through just the right action of $f$--if we require $\hat{\theta}^{(\sigma)}\hat{e}_{(\mu)}= \delta^\sigma_\mu$. But my understanding is that this restriction is not required. I believe (but am not certain) that this would end up being circular anyway, as the orthogonality relation is precisely imposed on dual spaces, and we are trying to prove that the gradient of $f$ is the one form that maps $T_p\to \mathbb{R}$. But then I don't see how this expression ends up equalling $\frac{df}{d\lambda}$.

• Consider that $v^{\mu} = dx^{\mu}/d\lambda$. Then it follows $v^{\mu} \partial_{\mu} f = df / d \lambda$, don't you agree? – OkThen Sep 9 '18 at 2:31
• Okay, fair.. but we still need to get the indices to match. That is, we have to contract $v^\mu \partial_\sigma f$. – InertialObserver Sep 9 '18 at 2:33
• Ok. I'm not claiming that I understood what you are asking, that's why I did't wrote an answer. I am responding to: "But then I don't see how this expression ends up equalling $d f /d \lambda$". You don't like $\theta^{\nu} ( e_{\mu} ) = \delta^{\nu}_{\mu}$ why? – OkThen Sep 9 '18 at 2:36
• Look, let us start with baby steps... $v = v^{\mu} e_{\mu}$ is a linear space. The dual space is what maps vectors to fields. Ok. But because your space is linear, the dual space will also inherit a linear structure. Find a basis on it, and call it $\theta^{\mu}$. Now, if you want specify complete action on the original vector space, all you have to do is specify how ithis $\theta$'s act on the basis $e$. Normalize it to $\theta^{\nu} (e_{\mu})= \delta^{\nu}_{\mu}$. This has nothing to do with orthogonality, there is no metric here. It is a construction. – OkThen Sep 9 '18 at 2:47
• I understand it all now. It was a logical error as well as a conceptual error on my part on what we were trying to actually show. Thanks! – InertialObserver Sep 9 '18 at 2:59