# I Differential forms and bases

#### kiuhnm

In the exercises on differential forms I often find expressions such as $$\omega = 3xz\;dx - 7y^2z\;dy + 2x^2y\;dz$$ but this is only correct if we're in "flat" space, right?
In general, a differential $1$-form associates a covector with each point of $M$. If we use some coordinates $(x^i)$ on an open set $U$ of $M$, then an expression like $3xz$ is correct and is valid for all points in $U$, but shouldn't $dx$ represent a different basis vector for each point of $U$? I guess if I write $3xz\;dx$ I'm assuming that $(dx)_p = (dx)_q$ for all $p,q\in U$.

Sorry for the many edits.

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#### fresh_42

Mentor
2018 Award
The (co-)vectors $dx,dy,\ldots$ are the basis vectors of the tangent space, which is flat. But $(dx)_p \neq (dx)_q$ in general. E.g. if $f(x)=x^2$ we get $df = 2x\,dx$ which is obviously different for different $x=p$. It's just that in a neighborhood $U$ of $p$ it is still close, however, different points yield different tangent spaces, even if they could be expressed in the same way, $2x\,dx$ in the example.

#### WWGD

Gold Member
In the exercises on differential forms I often find expressions such as $$\omega = 3xz\;dx - 7y^2z\;dy + 2x^2y\;dz$$ but this is only correct if we're in "flat" space, right?
In general, a differential $1$-form associates a covector with each point of $M$. If we use some coordinates $(x^i)$ on an open set $U$ of $M$, then an expression like $3xz$ is correct and is valid for all points in $U$, but shouldn't $dx$ represent a different basis vector for each point of $U$? I guess if I write $3xz\;dx$ I'm assuming that $(dx)_p = (dx)_q$ for all $p,q\in U$.

Sorry for the many edits.
Yes, if I understand you correctly, dx is toe localized to some point when evaluating it.

#### kiuhnm

The (co-)vectors $dx,dy,\ldots$ are the basis vectors of the tangent space, which is flat. But $(dx)_p \neq (dx)_q$ in general. E.g. if $f(x)=x^2$ we get $df = 2x\,dx$ which is obviously different for different $x=p$. It's just that in a neighborhood $U$ of $p$ it is still close, however, different points yield different tangent spaces, even if they could be expressed in the same way, $2x\,dx$ in the example.
I assume you meant cotangent space.

#### WWGD

Gold Member
I assume you meant cotangent space.
Yes, you are correct. And these are the duals to the standard del/delx basis of the tangent space.

#### fresh_42

Mentor
2018 Award
I assume you meant cotangent space.
To me these are all vector spaces and the "co" is simply what you do with them, resp. how you write it. Too much trouble to find out each time whether $v$ or $\langle v, .\rangle$ is meant. But you are right, the "form" in differential form indicates that it is a multilinear form.

#### fresh_42

Mentor
2018 Award
I assume you meant cotangent space.
I once had the fun and listed on how many ways a differentiation can be viewed. I came up with ten, and the word slope wasn't even among them.

#### kiuhnm

To me these are all vector spaces and the "co" is simply what you do with them, resp. how you write it. Too much trouble to find out each time whether $v$ or $\langle v, .\rangle$ is meant. But you are right, the "form" in differential form indicates that it is a multilinear form.
I understand that a covector is just a vector, but can we say that a cotangent space is just a tangent space? They're both vector spaces but are they both tangent to the manifold at a point? To me "tangent" means that it has to do with derivations, whereas cotangent means it's related to vanishing functions.

#### fresh_42

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2018 Award
I understand that a covector is just a vector, but can we say that a cotangent space is just a tangent space? They're both vector spaces but are they both tangent to the manifold at a point? To me "tangent" means that it has to do with derivations, whereas cotangent means it's related to vanishing functions.
No, a cotangent space is not a tangent space.

However, with the detour $V^* = W$ we can certainly find a manifold which $W$ is the tangent space for. But within the same setup I wouldn't confuse them.

In our example, $df_{1}= 2\cdot 1 \,dx$ and the actual tangent at $x=1$ is $y=2x-1$ so $"2"$ is the factor at $x$, the linear function $x \mapsto 2x\,.$ So cotangent has been correct.

But as I said, it depends on what we want to do with it: $2$ is a vector and $2\cdot ()$ a covector. At school it is a number which we call slope. And if you want to read the entire list ... well, it's in a language you don't know yet, as you've said, but if you want to get an impression: https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ pretty much at the beginning after the one-parameter group.

#### lavinia

Gold Member
Here is a formal distinction between tangent and cotangent spaces that may be of help.

If $F:M→N$ is a differentiable function and $X_{p}$ is a tangent vector at a point $p$ of $M$ then the differential of $F$ applied to $X_{p}$ $dF(X_{p})$ is a tangent vector at $F(p)$ in $N$.

Across the whole manifold $M$ one gets a map $F_{*}:TM→TN$ of the tangent bundle of $M$ into the tangent bundle of $N$.

For 1 forms one gets a map in the other direction. One has a map $F^{*}:TN^{*}→TM^{*}$ from the cotangent bundle of $N$ into the cotangent bundle of $M$. $F^{*}(ω)(X_{p})$ is defined to be $ω(dF(X_{p}))$.

Since the derivative is a linear map at each point and 1 forms are linear functionals $F^{*}$ and $F_{*}$ are both linear on each tangent/cotangent space.

Further if one has a composition of two functions $F:M→N$ and $G:N→S$ then $(G \circ F)_{*}= G_{*} \circ F_{*}$ and $(G \circ F)^{*}= F^{*} \circ G^{*}$
This is just the Chain Rule.

In Category language, the derivative is a covariant functor on tangent spaces and a contravaraint functor on cotangent spaces.

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#### WWGD

Gold Member
Still, in the finite-dimensional case, the two are equal, tho only naturally equal ( meaning isomorphism is independent of choice of basis) if the manifold has with it a non-degenerate quadratic form.

#### lavinia

Gold Member
Still, in the finite-dimensional case, the two are equal, tho only naturally equal ( meaning isomorphism is independent of choice of basis) if the manifold has with it a non-degenerate quadratic form.
One would not say that they are "naturally" equal since the isomorphism depends on the choice of quadratic form.

One would not say they "equal" but rather that they are isomorphic as vector spaces. But all vector spaces of the same finite dimension are isomorphic.

An example of a natural isomorphism is between a vector space to its double dual. If $v$ is a vector and $l$ is a linear functional then $v(l)=l(v)$. There is no need for a quadratic form or any other additional structure.

In the case of the tangent bundle and the bundle of 1 forms the isomorphism is more than just an isomorphism of each tangent and cotangent plane. It is also a bundle isomorphism.

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#### kiuhnm

Here is a formal distinction between tangent and cotangent spaces that may be of help.

If $F:M→N$ is a differentiable function and $X_{p}$ is a tangent vector at a point $p$ of $M$ then the differential of $F$ applied to $X_{p}$ $dF(X_{p})$ is a tangent vector at $F(p)$ in $N$.

Across the whole manifold $M$ one gets a map $F_{*}:TM→TN$ of the tangent bundle of $M$ into the tangent bundle of $N$.

For 1 forms one gets a map in the other direction. One has a map $F^{*}:TN^{*}→TM^{*}$ from the cotangent bundle of $N$ into the cotangent bundle of $M$. $F^{*}(ω)(X_{p})$ is defined to be $ω(dF(X_{p}))$.

Since the derivative is a linear map at each point and 1 forms are linear functionals $F^{*}$ and $F_{*}$ are both linear on each tangent/cotangent space.

Further if one has a composition of two functions $F:M→N$ and $G:N→S$ then $(G \circ F)_{*}= G_{*} \circ F_{*}$ and $(G \circ F)^{*}= F^{*} \circ G^{*}$
This is just the Chain Rule.

In Category language, the derivative is a covariant functor on tangent spaces and a contravaraint functor on cotangent spaces.
Basically, you have defined the tangent space and the cotangent space by the push-forward and pull-back induced by a map $F:M\to N.$ One can also note that the matrix associated with $F_*$ is just the Jacobian matrix of $F$ (which is more or less equivalent to your remark about the Chain Rule).
I don't know if I'll ever learn Category Theory, but it certainly gets straight to the point!

#### lavinia

Gold Member
Basically, you have defined the tangent space and the cotangent space by the push-forward and pull-back induced by a map $F:M\to N.$ One can also note that the matrix associated with $F_*$ is just the Jacobian matrix of $F$ (which is more or less equivalent to your remark about the Chain Rule).
I don't know if I'll ever learn Category Theory, but it certainly gets straight to the point!
Right except these properties are not meant to define the tangent and cotangent spaces but rather to illustrate their difference. You can certainly replace the push forward with the Jacobian matrix but this will only be a local representation with respect to charts on $M$ and $N$. Across the entire manifolds there may not be a global basis for the tangent space. For instance there is no global basis for the tangent space of the sphere.

By contrast, the push forward is is a mapping defined globally on the entire tangent bundle to $M$.

#### WWGD

Gold Member
One would not say that they are "naturally" equal since the isomorphism depends on the choice of quadratic form.

One would not say they "equal" but rather that they are isomorphic as vector spaces. But all vector spaces of the same finite dimension are isomorphic.

An example of a natural isomorphism is between a vector space to its double dual. If $v$ is a vector and $l$ is a linear functional then $v(l)=l(v)$. There is no need for a quadratic form or any other additional structure.

In the case of the tangent bundle and the bundle of 1 forms the isomorphism is more than just an isomorphism of each tangent and cotangent plane. It is also a bundle isomorphism.
I have always seen it used this way. Still, once a chice of quadratic form is given, the isomorphism is natural. Maybe one can find a different term for it, but you may even argue in some cases that the form comes naturally with a manifold .

#### lavinia

Gold Member
I have always seen it used this way. Still, once a chice of quadratic form is given, the isomorphism is natural. Maybe one can find a different term for it, but you may even argue in some cases that the form comes naturally with a manifold .
OK. But its not the terminology so much as the difference between the dual and the double dual that is the point. The double dual is natural in the sense that no other structure is needed to define the isomorphism. With the cotangent space more structure is needed for instance a quadratic form. This point about the naturally of the isomorphism to the double dual and the non-naturality of the isomorphism with the dual is a standard point in linear algebra. A manifold is not needed.

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#### WWGD

Gold Member
O\K. But its not the terminology so much as the difference between the dual and the double dual that is the point. The double dual is natural in the sense that no other structure is needed to define the isomorphism. With the cotangent space more structure is needed for instance a quadratic form. This point about the naturally of the isomorphism to the double dual and the non-naturality of the isomorphism with the dual is a standard point in linear algebra. A manifold is not needed.
I understand it is a general result and not just about manifolds, I just brought this up because it is the setting we ate discussion.

#### kiuhnm

Right except these properties are not meant to define the tangent and cotangent spaces but rather to illustrate their difference. You can certainly replace the push forward with the Jacobian matrix but this will only be a local representation with respect to charts on $M$ and $N$. Across the entire manifolds there may not be a global basis for the tangent space. For instance there is no global basis for the tangent space of the sphere.

By contrast, the push forward is is a mapping defined globally on the entire tangent bundle to $M$.
Speaking of push-forward, one book says that it's also called differential, but another book defines the differential differently: $df(X_p) = X_p(f)$. Which is it?
I also noticed that your definition of pull-back is somewhat different from mine. Your definition is $(f^\star w)(X_p) := w(f_\star X_p)$ i.e. you push-forward the vector, whereas my book's definition is simply $f^*w := w\circ f$. The push-forward is then defined through the pull-back.

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#### fresh_42

Mentor
2018 Award
$\varphi \, : \, M\longrightarrow N\,\,\, , \,\,\,\varphi_* \, : \, TM \rightarrow TN\,\,\, , \,\,\,\varphi^*\, : \,T^*N \longrightarrow T^*M$

Pushforward: $(\varphi_*(X_p))(f)=X_p(f \circ \varphi)\; , \; p \in M, X_p \in T_pM, f \in C^\infty(N)$

Possible notations: $\varphi_*(p,X)=\varphi’_p(X)=\varphi'(p)X=D \varphi_p(X) =D_p\varphi(X)=d\varphi_p(X)=d_p\varphi(X)=T_p\varphi(X)$ and the tangent vector can also be written with an additional $p$ if there is only one point under consideration, so $D \varphi_p(X_p)$ would be a double notation of the point.

Pullback: $(\varphi^*\nu)(p) = \nu(\varphi(p))\; , \;p \in M, \nu \in N^*$

for differential forms: $\varphi^*(d \omega) = d\varphi^*(\omega)$
for cotangents: $(\varphi^* \sigma)_p (X_p) :=\sigma_{\varphi(p)}(J_p(\varphi)(X_p)) =\sigma_{\varphi(p)}(\varphi_*(X_p)) = \sigma_{\varphi(p)}(D_p\varphi(X_p))\; , \;p\in M, X\in TM, \sigma \in T^*N$

#### lavinia

Gold Member
I also noticed that your definition of pull-back is somewhat different from mine. Your definition is $(f^\star w)(X_p) := w(f_\star X_p)$ i.e. you push-forward the vector, whereas my book's definition is simply $f^*w := w\circ f$. The push-forward is then defined through the pull-back.
$f^*w := w\circ f$ doesn't make any sense. The pull back of a 1 form is defined on the push forward of tangent vectors. Maybe you meant $f^*w := w\circ df$ or equivalently, $f^*w := w\circ f_{*}$

As fresh_42 pointed out, the key to the push forward is that if $c(t)$ is a curve in $M$ then $(f \circ c)(t)$ is a curve on $N$.

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#### kiuhnm

$f^*w := w\circ f$ doesn't make any sense.
It seems to me fresh_42 gave the same exact definition I'm using: $(\phi^* \nu)(p) = \nu(\phi(p)) = (\nu\circ\phi)(p)$.
His expression for differential forms is just a property of the $d$ operator, according to my book.

In $(f^*(w))(X_p) := w(f_* X_p)$ you do the pullback on $w$ by effectively pushing forward the vector $X_p$, but one can also just pull back $w$ itself by composing it with $f$: $(f^* w)(X_p) := w(f(X_p))$. That's my understanding at least.

BTW, I've just found out that we can also push forward and pull back tensorfields and if $f$ is diffeomorphic, we can even pullback by the inverse of the pushforward and vice versa!

#### fresh_42

Mentor
2018 Award
His expression for differential forms is just a property of the $d$ operator, according to my book.
It is, but it's quite catchy.
BTW, I've just found out that we can also push forward and pull back tensorfields ...
$(f^* \mathcal{F})_{p}(X_1,\ldots,X_k)=\mathcal{F}_{f(p)}(f_{*}(X_1 \otimes \ldots \otimes X_k)=\mathcal{F}_{f(p)}(d_pf(X_1),\ldots ,d_pf(X_k))$
... and if $f$ is diffeomorphic ...
$J_p(f)=f_*=d_pf \in GL(T_pM,T_{f(p)}N)$
It seems to me fresh_42 gave the same exact definition I'm using: $(\phi^* \nu)(p) = \nu(\phi(p)) = (\nu\circ\phi)(p).$
No, I did not. You wrote
$f^*w := w\circ f$
which does not make sense, as the left hand side applies to tangent vectors of $T_pM$, whereas the right hand side applies to points on $M.$

I wrote
$(\varphi^*\nu)(p) = \nu(\varphi(p))\; , \;p \in M, \nu \in N^*$
where $\varphi^*$ applies to linear forms on $N$ and thus has a linear form on $M$ as value and a scalar if evaluated at a certain point, which $\varphi$ still applies to.

@lavinia wrote
$F^{*}(ω)(X_{p})$ is defined to be $ω(dF(Xp))$
which applies to differential forms in general and corresponds to what I have said, too, about cotangents
for cotangents: $(\varphi^* \sigma)_p (X_p) :=\sigma_{\varphi(p)}(J_p(\varphi)(X_p))$
It is indeed a bit confusing, because there are so many different objects taking the role of variables: points, linear forms, differential forms, tangents, cotangents, differentials, multilinear forms, tensors, or whatever (e.g. sections).

Maybe you should really have a look on where I copied my formulas from:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/

#### lavinia

Gold Member
It seems to me fresh_42 gave the same exact definition I'm using: $(\phi^* \nu)(p) = \nu(\phi(p)) = (\nu\circ\phi)(p)$.

In $(f^*(w))(X_p) := w(f_* X_p)$ you do the pullback on $w$ by effectively pushing forward the vector $X_p$, but one can also just pull back $w$ itself by composing it with $f$: $(f^* w)(X_p) := w(f(X_p))$.
You must mean $(f^* w)(X_p) := w(f_{*}(X_p))$.

Otherwise it makes no sense.

#### kiuhnm

@fresh_42 I see it now. Thank you so much for your very detailed post! The book I'm reading does define the pullback of maps on manifolds. I got confused because it doesn't give an explicit formula for the pullback of forms. Instead, it says that the pullback can be extended to differential forms by introducing the following axioms: \begin{align*} f^*(\lambda+\mu) &= f^*\lambda + f^*\mu \\ f^*(\lambda\wedge\mu) &= f^*\lambda \wedge f^*\mu \\ d(f^*\lambda) &= f^*(d\lambda) \end{align*} Then, it writes the form $w$ as $w = \sum a_i dy^i$ and derives $$f^* w = \sum (a_i\circ f)\frac{\partial f^i}{\partial x^j} dx^j.$$ I understood the steps but I lost the full picture.
edit: In other words, I mistakenly thought that $f^* w$ was still defined as before and that the derivation above was just a more explicit way of writing the same thing. It's clear I wasn't connecting the dots...

#### kiuhnm

You must mean $(f^* w)(X_p) := w(f_{*}(X_p))$.

Otherwise it makes no sense.
Yes, I get it now. See my reply to @fresh_42. Thank you too!