Tensor field from a covariant derivative operator (affine connection)

[cianfa72]

TL;DR Summary

How to get a tensor field from a covariant derivative operator (affine connection)

Consider the expression

$$T(\omega, X, Y) := \omega \nabla_X Y$$
where $\omega,X,Y$ are a covector field and two vector fields respectively.

Is $T$ a (1,2) tensor field ? From my understanding the answer is negative. The point is that $T(\,. , \, . , \, .)$ is not $C^{\infty}$-linear in each of its three slots.

Nevertheless if one picks a point P on the manifold (e.g. spacetime) and plugs into $T$ slots the three smooth fields evaluated at P, then one gets a scalar. This because $\nabla_XY$ evaluated at P is a vector and its contraction with the covector field $\omega$ evaluated at P gives a scalar.

Does the above make sense ? Thanks.

 

[PeterDonis]

Consider the expression

$$T(\omega, X, Y) := \omega \nabla_X Y$$
where $\omega,X,Y$ are a covector field and two vector fields respectively.

What does this expression mean? I know what $\nabla_X Y$ means, but I don't know what it means to put $\omega$ on its left.

 

[cianfa72]

What does this expression mean? I know what $\nabla_X Y$ means, but I don't know what it means to put $\omega$ on its left.

It is basically the contraction of the covector $\omega$ evaluated at point P with $\nabla_X Y$ evaluated at P.

 

[PeterDonis]

It is basically the contraction of the covector $\omega$ evaluated at point P with $\nabla_X Y$ evaluated at P.

A contraction would normally be written as something like $\omega \cdot \nabla_X Y$. And such a contraction would be a scalar: $\nabla_X Y$ is a vector field (because $Y$ is a vector field and $\nabla_X$ is a scalar operator--or, if you like, you can think of it as taking $\nabla Y$, which is a (1, 1) tensor field, and contracting it with the vector field $X$ to produce a (1, 0) tensor field, i.e., a vector), and contracting a vector with a covector gives a scalar.

 

[PeterDonis]

@cianfa72 , this is an example where Wald's abstract index notation really helps. $\omega$ is a covector, so it is written $\omega_a$. $\nabla_X Y$ is written $X^b \nabla_b Y^a$, which makes it obvious that it's a vector. Then $\omega \cdot \nabla_X Y$ is written $\omega_a X^b \nabla_b Y^a$, which makes it obvious that it's a scalar (no free indexes are left).

 

[PeterDonis]

The point is that $T(\,. , \, . , \, .)$ is not $C^{\infty}$-linear in each of its three slots.

This question as you phrase it is moot because, as my previous posts have shown, you have incorrectly identified the actual tensor object that your expression constructs.

However, once we look at the correct object, we can see that it is, as I have said, a scalar, and the objects of which that scalar is composed by contraction are all tensors. There is nothing that isn't a tensor anywhere. So I don't know where your statement that there is something that isn't $C^\infty$ is coming from.

 

[Yuras]

you have incorrectly identified the actual tensor object that your expression constructs

I think they are operating on abstract tensors (aka multilinear maps). So $T$ is indeed a (1,2)-rank tensor since it takes a covector and two vectors and produces a number.
Though I don't understand why it won't be a tensor field as well. Probably some nuance of the rigorous definition of the tensor field.

 

[martinbn]

TL;DR Summary: How to get a tensor field from a covariant derivative operator (affine connection)

Consider the expression

$$T(\omega, X, Y) := \omega \nabla_X Y$$
where $\omega,X,Y$ are a covector field and two vector fields respectively.

Not to cause confusion you should write it as $T(\omega, X, Y) := \omega (\nabla_X Y)$

Is $T$ a (1,2) tensor field ? From my understanding the answer is negative. The point is that $T(\,. , \, . , \, .)$ is not $C^{\infty}$-linear in each of its three slots.

This is correct, $T$ is not a tensor field. Already $\nabla_X Y$ this is not a tensor field viewed as map of both $X$ and $Y$.

Nevertheless if one picks a point P on the manifold (e.g. spacetime) and plugs into $T$ slots the three smooth fields evaluated at P, then one gets a scalar. This because $\nabla_XY$ evaluated at P is a vector and its contraction with the covector field $\omega$ evaluated at P gives a scalar.

Does the above make sense ? Thanks.

 

[Yuras]

Though I don't understand why it won't be a tensor field as well. Probably some nuance of the rigorous definition of the tensor field.

Ah, that's because the coefficients $a_1$ and $a_2$ in $T(\omega, X, a_1 Y_1+a_2 Y_2)$ might depend on the point of the manifold, so we get extra terms from differentiation. Easy to overlook :) So yes, it's not a tensor field.

 

[martinbn]

Not always an acceptable source, but for this I think it is ok.

https://en.wikipedia.org/wiki/Tensor_field#Tensor_fields_as_multilinear_forms

 

[cianfa72]

This is correct, $T$ is not a tensor field. Already $\nabla_X Y$ this is not a tensor field viewed as map of both $X$ and $Y$.

You mean $\nabla$ viewed as an operator/map with two slots eating vector fields. It is not $C^{\infty}$-linear in the second slot (the slot that eats the vector field $Y$).

Btw, the map that "implement" the contraction of a covector with a vector is itself a (1,1) tensor since it a bi-linear map into scalars.

 

[PeterDonis]

the map that "implement" the contraction of a covector with a vector is itself a (1,1) tensor since it a bi-linear map into scalars.

I'm not sure this is correct as you state it. The correct way of stating it, as I understand it, is that a covector itself is a linear map from vectors to scalars. There is no additional map that needs to be "implemented" to do contraction of a covector with a vector; the covector itself is the map.

 

[cianfa72]

I'm not sure this is correct as you state it. The correct way of stating it, as I understand it, is that a covector itself is a linear map from vectors to scalars.

Yes, that's true. However consider the map that takes in its two slots a covector and a vector and returns their contraction. It is bi-linear then, by definition, it is a (1,1) tensor.

 

[PeterDonis]

consider the map that takes in its two slots a covector and a vector and returns their contraction.

There is no such map separately from the covector itself; the covector is this map. That's my point.

A (1, 1) tensor is a map that takes a covector and a vector and returns a number that is not just the contraction of the two.

 

[Orodruin]

Yes, that's true. However consider the map that takes in its two slots a covector and a vector and returns their contraction. It is bi-linear then, by definition, it is a (1,1) tensor.

That’s just the identity map on covectors. (Remember that a (1,1) tensor may also be viewed as a map from covectors to covectors)

There is no such map separately from the covector itself; the covector is this map. That's my point.

Hence why the (1,1) map (it does exist!) is the identity map from covectors to covectors. (Or from vectors to vectors for that matter.)

 

[Orodruin]

Ah, that's because the coefficients $a_1$ and $a_2$ in $T(\omega, X, a_1 Y_1+a_2 Y_2)$ might depend on the point of the manifold, so we get extra terms from differentiation. Easy to overlook :) So yes, it's not a tensor field.

This is why we need the asymmetry in the definitions of torsion and Riemann curvature (and even then we are left with a term depending on the vector commutator …) for them to define tensors.

 

[PeterDonis]

That’s just the identity map on covectors.

Ah, I see. That must be what I was groping towards in my previous posts.

 

[Orodruin]

Ah, I see. That must be what I was groping towards in my previous posts.

I have an unfair advantage: I have been using it as an example of a (1,1) tensor in an introductory course for the better part of a decade now …

 

[cianfa72]

That’s just the identity map on covectors. (Remember that a (1,1) tensor may also be viewed as a map from covectors to covectors).

Just to fix ideas consider in $\mathbb R^2$ as vector space the following: $$[e_1 \otimes e^1 + e_2 \otimes e^2] (\, ., \, .)$$
where $\{ e_i \}$ is a basis and $\{ e^i\}$ the associated dual basis. It is a (1,1) tensor that realizes the contraction of a covector (first slot) with a vector (second slot).

 

[Orodruin]

Just to fix ideas consider in $\mathbb R^2$ as vector space the following: $$[e_1 \otimes e^1 + e_2 \otimes e^2] (\, ., \, .)$$
where $\{ e_i \}$ is a basis and $\{ e^i\}$ the associated dual basis. It is a (1,1) tensor that realizes the contraction of a covector (first slot) with a vector (second slot).

Yes, although as I said it is a rather trivial map as it is just the identity map on covectors. Calling the map $I$, it holds trivially that $I(\omega, V) = \omega(V)$. You do not really need $I$ to define a contraction, the contraction is perfectly well defined by having $\omega$ act on $V$.

 

[Yuras]

You do not really need to define a contraction

Hold on. I'm probably missing something obvious here, but... Aren't you talking about different things? Your identity map is a map $V\oplus V^\star\to F$, while @cianfa72 seems to be talking about the canonical pairing, which is $V\otimes V^\star\to F$ (there $V$ is a vector space over field $F$). You can get the canonical map from the identity map by linear extension.
(I can't say I understand this topic well enough, so expect nonsense)
How you define contraction without the canonical map? E.g. you have (1,1) tensor $X$ and you want to contract it by indices. It's tempting to write $X(\omega, v)=\omega(v)$, but you have neither $\omega$, nor $v$, you actually need to construct a trace. Am I missing something again?

 

[Orodruin]

Your identity map is a map $V\oplus V^\star\to F$

No, it isn't. The identity map $I$ on $V^*$ is a map $V^* \to V^*$ such that $\omega \to I(\omega) = \omega$, from which there is a 1-to-1 correspondence to maps $V\otimes V^* \to F$.

How you define contraction without the canonical map? E.g. you have (1,1) tensor $X$ and you want to contract it by indices. It's tempting to write $X(\omega, v)=\omega(v)$, but you have neither $\omega$, nor $v$, you actually need to construct a trace. Am I missing something again?

@cianfa72 is not contracting the indices of $X$ (which would be taking the trace), they are introducing a map $X$ such that $X(\omega, V) = \omega(V)$ for all $\omega$ and $V$.

 

[Orodruin]

A short note on how maps $V^* \to V^*$ are identified with maps $V \otimes V^* \to F$:

Assuming $T: V \otimes V^* \to F$ is a multilinear map then for any basis $E_a$ with corresponding dual basis $E^a$ (such that $E^a (E_b) = \delta^a_b$) and for any $\omega\in V^*$ we can construct $T'(\omega) = T(E_a, \omega) E^a$, which is in $V^*$ as $E^a \in V^*$ and $T(E_a,\omega)$ is a number. Since $T'(\omega)$ is linear in $\omega$ it follows that $T'$ is a linear map $V^* \to V^*$ uniquely defined by $T$.

Assuming $T': V^* \to V^*$ is a linear map then for any $\omega \in V^*$ it holds that $T'(\omega) \in V^*$ by definition. We can then define $T: V \otimes V^*$ by $T(X,\omega) = T(\omega)(X)$, which is a linear map $V \otimes V^*$ uniquely defined by $T'$.

It is easy to verify that the above processes are each others' inverses as well as to generalise to any type of tensor and moving $V$s and $V^*$s from the domain to the codomain of the multilinear map.

 

[Yuras]

No, it isn't. The identity map $I$ on $V^*$ is a map $V^* \to V^*$ such that $\omega \to I(\omega) = \omega$, from which there is a 1-to-1 correspondence to maps $V\otimes V^* \to F$.

IIRC the statement is that there is a natural isomorphism between multilinear maps $V\oplus V^\star\to F$ and linear maps $V\otimes V^\star\to F$. You said it can be constructed, which doesn't follow from existence, hence the question.

@cianfa72 is not contracting the indices of $X$ (which would be taking the trace), they are introducing a map $X$ such that $X(\omega, V) = \omega(V)$ for all $\omega$ and $V$.

Hmm, yeah. Than do you by any chance know whether there is a way to construct the more general contraction?

 

[Orodruin]

IIRC the statement is that there is a natural isomorphism between multilinear maps $V\oplus V^\star\to F$ and linear maps $V\otimes V^\star\to F$. You said it can be constructed, which doesn't follow from existence, hence the question.

The statement is that there is a natural isomorphism between multilinear maps $V \otimes V^* \to F$ and linear maps $V^* \to V^*$. See post #25 where I constructed the isomorphism explicitly.

Hmm, yeah. Than do you by any chance know whether there is a way to construct the more general contraction?

Yes! Take any pair of bases $E_a$ and $E^a$ for $V$ and $V^*$, respectively, such that $E^a(E_b) = \delta^a_b$. Then the trace of $X: V \otimes V^* \to F$ is defined as $$
\operatorname{Tr}(X) = X(E_a, E^a)
$$ (Einstein summation implied) It is relatively easy to check that this is indeed independent of the pair of bases.

Other contractions may be constructed similarly.

 

[Yuras]

The statement is that there is a natural isomorphism between multilinear maps $V \otimes V^* \to F$ and linear maps $V^* \to V^*$.

Ah, so you are just a bit sloppy in your notation, which confuses me. Technically $V \otimes V^* \to F$ can't be multilinear (well, it can, but only trivially) since it takes only one argument. I mean, these maps are naturally isomorphic, so usually there is no need to differentiate between them, but right now it's kinda the point :) It's hard to understand whether you mean isomorphism between multilinear maps $V \oplus V^* \to F$ and linear maps $V^* \to V^*$, or between linear maps $V \otimes V^* \to F$ and linear maps $V^* \to V^*$

See post #25 where I constructed the isomorphism explicitly.

There you assume existence of $E_a$. I mean, it surely exists, but the question is about constructing it.

Even if we take it as a proof of existence, I'm still a bit suspicious because I think it should fail in infinite dimentional case, and I don't see where exactly it fails. Though I'm not certain in this claim. ADDED: nevermind, the proof fails for infinite dimentions because you are implicitly assuming $V^{**}=V$

 

[Orodruin]

There you assume existence of Ea. I mean, it surely exists, but the question is about constructing it.

You can take any linearly independent basis for $V$. Starting from that and a linearly independent basis for $V^*$ you can use the equivalent of the Gram-Schmidt process to find the correct basis for $V^*$.

You could also just define the basis on $V^*$ by its action on the basis in $V$.

I think it should fail in infinite dimentional case,

This being the relativity forum and not the differential geometry forum I think it is fine to say that we consider finite vector spaces. Generally n=2 or 4 will do the trick.

I think I was also confused by your notation. You are likely thinking of a different 1-1 correspondence, namely that of $t: V\times V^* \to F$ to $T: V\otimes V^* \to F$ given by $t(v,\omega) = T(v\otimes \omega)$. This is not what I was talking about at all. I was talking about the correspondence of maps $V^* \to V^*$ to maps $V\times V^* \to F$.

 

[Yuras]

I think I was also confused by your notation.

Oops, my bad. Indeed, I should have been using direct product $\times$ instead of a direct sum $\otimes$. I somehow failed to learn the difference, and confuse them a lot. Sorry for that. I guess the whole thread now makes no sense because of multiple layers of confusion :)

You are likely thinking of a different 1-1 correspondence, namely that of $t: V\times V^* \to F$ to $T: V\otimes V^* \to F$ given by $t(v,\omega) = T(v\otimes \omega)$.

Well, I was interested in tensor contraction only (the general one, not restricted to $\omega(v)$), all these maps appeared only because I tried to understand what you are talking about.
I actually though that in #20 you are talking about things like this (which I fail to comprehend to he honest)

This is not what I was talking about at all. I was talking about the correspondence of maps $V^* \to V^*$ to maps $V\times V^* \to F$.

Now that I understand which isomorphism you are talking about... I think this one can be constructed without involving basis. I.e. given $f :V\times V^* \to F$ we can construct $g:V^* \to V^*$ like this: $g(\omega) =v \mapsto f(v,\omega)$. And the other direction: $f(v, \omega)=g(\omega)(v)$

 

[cianfa72]

Assuming $T': V^* \to V^*$ is a linear map then for any $\omega \in V^*$ it holds that $T'(\omega) \in V^*$ by definition. We can then define $T: V \otimes V^*$ by $T(X,\omega) = T(\omega)(X)$, which is a linear map $V \otimes V^*$ uniquely defined by $T'$.

The latter should actually be $T(X,\omega) := T'(\omega)(X)$.

Coming back to the OP, on the other hand the quantity $$(\nabla_XY)$$ where $X,Y$ are smooth vector fields is a (smooth) vector field since it acts as a $C^{\infty}$ linear map on a covector field. Indeed its action on the covector field $\omega$ is defined as $$(\nabla_XY)(\omega)=\omega (\nabla_XY)$$ by using the canonical isomorphism/identification of a vector with the bi-dual. The above is $C^{\infty}$ linear map since $$(\nabla_XY)(f\omega)=f\omega (\nabla_XY)=f(\nabla_XY)(\omega)$$ for any $f \in C^{\infty}$.