Tangent vectors as directional derivatives

"Don't panic!"

Sorry to bring up a previous question, but why would it be incorrect to do a proof of Hadamard's lemma using the following approach?!

Let $\gamma :[0,1]\rightarrow\mathbb{R}^{n}$ be a curve in $\mathbb{R}^{n}$ such that $\gamma (0)=a\in\mathbb{R}^{n}$ and $\gamma (1)=x\in\mathbb{R}^{n}$, hence a fixed value of $t\in [0,1]$ identifies a point in $\mathbb{R}^{n}$ by $\gamma (t)$. The curve $\gamma$ is such that it passes through the fixed point $a\in\mathbb{R}^{n}$ and some arbitrary point $x\in\mathbb{R}^{n}$. The coordinate representation of the curve is given by $$\gamma (t)=\left(x^{1}(t),\ldots,x^{n}(t)\right)=\gamma (0)+t\left(\gamma (1)-\gamma (0)\right)=a+t\left(x-a\right)$$ where $x^{\mu}:[0,1]\rightarrow\mathbb{R}$ are the coordinate functions that specify the coordinates of the curve $\gamma$ in $\mathbb{R}^{n}$ for each value of $t\in [0,1]$. We thus define the coordinate functions $x^{\mu}$ such that $$x^{\mu}(t)=a^{\mu}+t\left(x^{\mu}-a^{\mu}\right)$$ where $a^{\mu}, x^{\mu}\in\mathbb{R}$.

Now consider the function $h:[0,1]\rightarrow\mathbb{R}$, defined such that $$h(t)=(F\circ\gamma)(t)=F(\gamma(t))=F(a+t(x-a))$$ where $F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is $C^{\infty}$. It follows that $$\frac{dh}{dt}=h'(t)=\frac{d}{dt}\left[(F\circ\gamma)(t)\right]=\frac{d}{dt}\left[F\left((x^{1}(t),\ldots,x^{n}(t))\right)\right]\\ \qquad=\sum_{\mu =1}^{n}\frac{\partial F}{\partial x^{\mu}}\frac{dx^{\mu}}{dt}=\sum_{\mu =1}^{n}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]\left(x^{\mu}-a^{\mu}\right).$$ Observe that $$h(1)-h(0)=F(x)-F(a)=\int_{0}^{1}h'(t)dt=\int_{0}^{1}\sum_{\mu =1}^{n}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]\left(x^{\mu}-a^{\mu}\right)dt =\sum_{\mu =1}^{n}\left(x^{\mu}-a^{\mu}\right)\int_{0}^{1}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]dt \\ =\sum_{\mu =1}^{n}\left(x^{\mu}-a^{\mu}\right)H_{\mu}(x)$$ where $H_{\mu}(x)= \int_{0}^{1}\frac{\partial}{\partial x^{\mu}}\left[F(a+t(x-a))\right]dt$.
Note that $F$ is $C^{\infty}$ and so it follows that $H_{\mu}$ are also $C^{\infty}$.

Finally, taking derivatives of both sides with respect to $x^{\mu}$ we have that $$\frac{\partial F}{\partial x^{\mu}}=H_{\mu}(x)+\left(x^{\nu}-a^{\nu}\right)\frac{\partial H_{\nu}}{\partial x^{\mu}}$$ and so evaluating this at $x=a$ gives $$\frac{\partial F}{\partial x^{\mu}}\bigg\vert_{x=a}=H_{\mu}(a)$$ as claimed.

Fredrik

Staff Emeritus
Gold Member
I don't think there's anything wrong with it. I think the following version is slightly simpler though. I would write
\begin{align}
F(x)-F(a) &=\int_0^1 \frac{d}{dt}F\big(a+t(x-a)\big) dt =\int_0^1 \left(\sum_{i=1}^n F_{,i}\big(a+t(x-a)\big) (x-a)_i \right)dt\\
&=\sum_{i=1}^n (x-a)_i \int_0^1 F_{,i}(a+t(x-a)) dt,
\end{align} then define $H_i$ by
$$H_i(x)=\int_0^1 F_{,i}(a+t(x-a)) dt$$ for all $x$, and then note that
$$H_i(a)=\int_0^1 F_{,i}(a) dt =F_{,i}(a)\int_0^1 dt =F_{,i}(a).$$ This is essentially the same thing you did in a different notation.

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"Don't panic!"

Thanks for taking a look at it. I know I was quite explicit with all the steps, but I just wanted to check that I was doing all correctly really. Was the first part of, with defining the curve $\gamma$, etc?!

Fredrik

Staff Emeritus
Gold Member
That part is strangely worded. Your "let $\gamma$" statement makes it sound like $\gamma$ is an arbitrary curve such that $\gamma(0)=a$ and $\gamma(1)=x$, and later you're using that $\gamma(t)=a+t(x-a)$, so it looks like you're making the claim that every curve $\gamma$ such that $\gamma(0)=a$ and $\gamma(1)=x$ is a straight line. It would be better to start with something like this: Let $\gamma:\mathbb R\to\mathbb R^n$ be the curve defined by $\gamma(t)=a+t(x-a)$ for all $t\in\mathbb R$. (The domain of this curve doesn't have to be $\mathbb R$, but if you want to say that the curve is smooth on [0,1], the domain must be a set that contains an open set that contains [0,1]).

"Don't panic!"

It would be better to start with something like this: Let γ:R→Rn\gamma:\mathbb R\to\mathbb R^n be the curve defined by γ(t)=a+t(x−a)\gamma(t)=a+t(x-a) for all t∈Rt\in\mathbb R.
So would it be better to start with this and then require that $\gamma$ is smooth in the interval $[0,1]$, and then proceed with the rest of the proof as I have in the previous post (apart from the change in notation as you suggested)?

Fredrik

Staff Emeritus
Gold Member
If you define $\gamma:\mathbb R\to\mathbb R^n$ the way I did, you don't have to require it to be smooth. It will simply be smooth.

The notation is a matter of taste.

"Don't panic!"

Ok, great. Thanks for your help :)

lavinia

Gold Member
Would this be correct though?

This is what I was trying to explain to my friend, but he when I said that each $t\in (a, b)$ maps to a point on the manifold I couldn't convince him that $t$ doesn't define a 1-dimensional coordinate system, I think he was thinking in terms of classical cases (in Euclidean space), but I tried to explain that the case is the same there as well, as one can always parameterise a curve in Euclidean space such that each value of $t$ corresponds to a set of coordinate values, but the parameter $t$ itself isn't considered as a coordinate.
A curve does not define a coordinate chart on a 1-dimensional sumbamnifold for a couple of reasons.

- A coordinate chart on a 1 dimensional manifold maps an open neighborhood on the manifold into R. But a curve maps an open neighborhood in R into the manifold. This is called a " parameterization" of a neighborhood if it is invertible and its inverse is smooth.

- But for an arbitrary smooth curve, its inverse may not even be differentiable. For instance inverse of the curve $t-> t^3$ is not differentiable at 0.

- For a submanifold of a larger dimensional manifold, the a coordinate chart on the submanifold alone is not considered to be a coordinate chart on the submanifold unless it can be smoothly extended to an open neighborhood in the larger dimensional manifold as well. In general, a mapping of any subset of a smooth manifold is called smooth if it can be smoothly extended to an open neighborhood in the ambient manifold.

So a submanifold is defined as a subset so that around each point there is an open neighborhood in the subspace topology that is homeomorphic to an open set in euclidean space and such a homeomorphism can be chosen so that it and its inverse are smooth. This means that the homeomorphism must be smoothly extendable to an open set in the larger manifold.

I was trying to rationalise with him why the definition of a tangent vector using this approach is intrinsically coordinate independent? Is what I put correct, or is it more that as it is defined as an equivalence class of curves and therefore not dependent on any one particular curve it is independent of any coordinate system introduced when specifying the form of a particular curve?!
Your expression "intrinsically coordinate independent" seems vague to me. Can you define it more precisely?

While in my mind, intrinsic can have more than one meaning, one idea is that the calculation of the quantity always gives the same answer in any coordinate system. To me this is what "coordinate independent" means. It does not mean that you have to define it without coordinate systems.

While it is true that defining tangent vectors as equivalence classes of curves appears not to use coordinates, one is assuming the idea of differentiability and differentiable can not be defined without reference to coordinate systems. How do you know that a function on a manifold is differentiable? When its composition with a parameterization is a differentiable function on Euclidean space. That requires the idea of coordinate systems.

- If one does use coordinate systems then for each parameter neighborhood of a point, one can define the directional derivative of a function by composing it with the parameter mapping. This has the great advantage of immediately showing that the differential of the function is a linear map.

If you do it this way, you need to somehow identify directions in different parameterizations. This is done using the Chain Rule. So tangent vectors are thought of directions in different parameterizations that are pasted together - identified. This shows that tangent vectors are intrinsic.

It also shows that tangent vectors form a vector space since the differentials of coordinate transformations are linear maps.

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"Don't panic!"

- A coordinate chart on a 1 dimensional manifold maps an open neighborhood on the manifold into R. But a curve maps an open neighborhood in R into the manifold. This is called a " parameterization" of a neighborhood if it is invertible and its inverse is smooth.
My (perhaps incorrect?!) intuition behind this was that a curve is a map that assigns a real value $t$ to each point $p$ on a manifold $M$, such that the points trace out a curve on $M$, however, it does not localise the position of each point that it maps to on $M$; this requires the introduction of a local coordinate chart such that we can describe the positions of the points of the points on $M$ in $\mathbb{R}^{n}$.

Your expression "intrinsically coordinate independent" seems vague to me. Can you define it more precisely?
By this I basically meant what you have said, i.e. "the calculation of the quantity always gives the same answer in any coordinate system", but you're right, my wording was a little vague, apologies for that.

If you do it this way, you need to somehow identify directions in different parameterizations. This is done using the Chain Rule. So tangent vectors are thought of directions in different parameterizations that are pasted together - identified.
By this do you is it mean that the curves in each equivalence class identify a particular direction along the manifold (at some point on the manifold) thus providing a notion of direction to the tangent vector that each equivalence class is identified with?!

lavinia

Gold Member
My (perhaps incorrect?!) intuition behind this was that a curve is a map that assigns a real value $t$ to each point $p$ on a manifold $M$, such that the points trace out a curve on $M$, however, it does not localise the position of each point that it maps to on $M$; this requires the introduction of a local coordinate chart such that we can describe the positions of the points of the points on $M$ in $\mathbb{R}^{n}$.
I have not seen it defined this way. I think of a curve as the path of a particle.

By this do you is it mean that the curves in each equivalence class identify a particular direction along the manifold (at some point on the manifold) thus providing a notion of direction to the tangent vector that each equivalence class is identified with?!
Well if you mean by direction a vector.

Given a parameterization,φ, of a domain on a manifold, the composed function $F(x) = f(φ(x))$ can be differentiated with respect to a vector,h, by taking the limit of the Newton quotient

(F( x + th) - F(x))/t.

(By directional derivative I mean this rather than restricting h to be just a unit vector.)

This definition works with respect to any parameterization but one needs to compare vectors in two different parameterizations in order to say when the vectors are the same
This is done with coordinate transformations. The differential of a coordinate transformation will map some vector,v, in another parameterization to the vector,h,. v and h are considered to be the same in the tangent space to the manifold.

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"Don't panic!"

I've become confused in studying differential geometry as to whether direction is an intrinsic property of vectors or not?! My understanding from studying abstract vector spaces is that it isn't (there is no reference to a notion of direction or magnitude of a vector in the vector space axioms) and that one can only gain a notion of magnitude and direction when one introduces a norm and inner product for a given space?!

Fredrik

Staff Emeritus
Gold Member
I've become confused in studying differential geometry as to whether direction is an intrinsic property of vectors or not?! My understanding from studying abstract vector spaces is that it isn't (there is no reference to a notion of direction or magnitude of a vector in the vector space axioms) and that one can only gain a notion of magnitude and direction when one introduces a norm and inner product for a given space?!
Depends on what you mean by "direction". For any non-zero vector $x$ I think it's very natural to think of a curve of the form $t\mapsto tx$ as singling out a direction in that vector space. And since there's one such curve for each non-zero vector, we might as well say that each non-zero vector identifies a direction.

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