Explore Carroll's Theory on Dual Space and Real Numbers

[dontknow]

"The dual space is the space of all linear maps from the original vector space to the real numbers." Spacetime and Geometry by Carroll.
Dual space can be anything that maps a vector space (including matrix and all other vector spaces) to real numbers.
So why do we picked only a vector as a linear map? ( it can be a matrix for field tensor but is there an example other than vectors for a "linear" map from vector space to real number).

 

[Ibix]

I'm not sure I understand this question. Sure you can map higher rank tensors to scalars using tensors of the same rank, but not to map vectors to scalars.

Vectors and covectors are of interest because they are the simplest objects that live in the tangent and cotangent spaces, and once you've established their basis and transformation rules you can use these for higher rank tensors.

Tensors aren't matrices, by the way. You can write rank 2 tensors as matrices, but they have additional structure that isn't expressible that way (co- and contravariance) and don't obey the same multiplication rules in general.

 

[dontknow]

I'm not sure I understand this question. Sure you can map higher rank tensors to scalars using tensors of the same rank, but not to map vectors to scalars.

Vectors and covectors are of interest because they are the simplest objects that live in the tangent and cotangent spaces, and once you've established the basis and transformation rules you can use these for higher rank tensors.

Tensors aren't matrices, by the way. You can write rank 2 tensors as matrices, but they have additional structure that isn't expressible that way (co- and contravariance) and don't obey the same multiplication rules in general.

"Sure you can map higher rank tensors to scalars using tensors of the same rank, but not to map vectors to scalars." Thanks.
Trace is a linear mapping. So is it going to be in the cotangent space?

 

[Ibix]

Trace is a linear mapping. So is it going to be in the cotangent space?

Taking the trace requires a tensor with one basis vector and one basis covector, which you then contract over. So trace is an operation involving things in the tangent and cotangent spaces. It also only returns a scalar for rank 2 tensors. Carroll is looking for something general (and which turns out to underlie the trace operation).

 

[PeroK]

"The dual space is the space of all linear maps from the original vector space to the real numbers." Spacetime and Geometry by Carroll.
Dual space can be anything that maps a vector space (including matrix and all other vector spaces) to real numbers.
So why do we picked only a vector as a linear map? ( it can be a matrix for field tensor but is there an example other than vectors for a "linear" map from vector space to real number).

Take any linear map $L$ on a vector space $V$ and any basis $\{e_n\}$ for $V$. Now, consider the mapping of $L$ on $\{e_n\}$:
$$L(e_n) = a_n$$
For some set of real numbers $\{a_n\}$.

This defines the mapping $L$ for the whole of $V$, as linearity ensures that if $v = v^ne_n$, then:
$$L(v) = v^na_n$$
Now, if we associate the set of numbers $\{a_n\}$ with $L$ the we have a representation of $L$ as a vector with the same dimension as $V$. And we clearly have a one-to-one correspondence between linear maps and a dual vector space.

The next step, of course, is to identify a corresponding basis for the dual vectors. We can define a linear map:
$$w^m(e_n) = \delta^m_n$$
And, after a little bit of gentle algebra, we see that:
$$L = a_mw^m$$
That gives us a corresponding set of basis dual vectors and te components of $L$ in that basis. And, in particular, there can be no other linear maps.

Note, if you are reading Carroll you may be interested in watching these MIT lectures, which largely follow Carroll's notes and may illuminate the material somewhat:

https://ocw.mit.edu/courses/physics/8-962-general-relativity-spring-2020/video-lectures/

 

[martinbn]

Just to add that the elements of any vector space are usually called vectors, no matter what the vector space is. It can be the space of tangent vectors at a point on a manifold. It can be the space of matrices, the space of functions, tensors and so on. Then the dual space is the space of linear maps from the given space to the reals.

 

[PeroK]

Just to add that the elements of any vector space are usually called vectors, no matter what the vector space is. It can be the space of tangent vectors at a point on a manifold. It can be the space of matrices, the space of functions, tensors and so on. Then the dual space is the space of linear maps from the given space to the reals.

The convention is physics, however, is that a vector is a more specific term. In fact, a vector would be described as a $(1, 0)$ tensor. So, in GR, it is tensor that is considered the generic term, with a vector being a special case.

 

[martinbn]

The convention is physics, however, is that a vector is a more specific term. In fact, a vector would be described as a $(1, 0)$ tensor. So, in GR, it is tensor that is considered the generic term, with a vector being a special case.

In the first post he writes:

Dual space can be anything that maps a vector space (including matrix and all other vector spaces) to real numbers.

That's what my comment addresses.

 

[Dale]

So why do we picked only a vector as a linear map?

I don’t understand your question. In the quote you cited the dual space is also a linear map:

"The dual space is the space of all linear maps from the original vector space to the real numbers."

 

[PeroK]

In the first post he writes:

That's what my comment addresses.

That would explain the OP's confusion. The OP is reading Carroll, who reserves vector for a quantity with specific physical and transformational properties. And, especially the term four-vector.

The generic linear algebra terminology that matrices, tensors and everything else also form vector spaces is misleading. That's not the terminology that Carroll employs: a matrix or tensor is definitely not a vector.

 

[vanhees71]

Well, that's sometimes indeed confusing. Often in physics they call vector components and tensor components vectors and tensors, and these components have of course transformation properties under changes of the basis. Concerning vectors and dual vectors (aka co-vectors) the logics is

Plain vector spaces

There's a set called Vectors obeying the usual rules for vector addition and multiplication with scalars (here I assume the scalars are the real numbers). I assume that the vector space is finite-dimensional.

Then you can define linear forms, which are linear maps from the vector space to the real numbers.

You can now show that by defining a basis $\vec{b}_j$, you can write every vector uniquely as $\vec{v}=v^j \vec{b}_j$ (Einstein summation convention used). Then for a linear form you have
$$\underline{L}(\vec{v})=\underline{L}(v^j \vec{b}_j)=v^j \underline{L}(\vec{b}_j).$$
This means the linear form is completely determined by the numbers $L_j=L(\vec{b}_j)$ and this means the set of linear forms on $V$ build a vector space, which is called the dual vector space, $V^*$.

Now you can define the co-basis $\underline{b}^k$ of the dual space as those linear forms which map the basis vectors to $\underline{b}^k(\vec{b}_j)=\delta_j^k$. Then $\underline{L}=L_k \underline{b}^k$.

Now you can work out the behavior of the components under basis transformations (implying also how the corresponding dual basis transforms) by noting that the vectors and dual vectors are independent of the choice of the basis: $\vec{v}=v^j \vec{b}_j = v^{\prime j} \vec{b}_j'$ and $\underline{L}=L_k \underline{b}^k = L_k' \underline{b}^{\prime k}$.

Though $V$ and $V^*$ are both real vector spaces of the same (finite) dimension and thus they are isomorphic vector spaces, but you cannot give a basis-independent mapping between these spaces. This only changes when considering

(Pseudo-)Euclidean vector spaces

A (pseudo-)Euclidean vector space has in addition a non-degenerate symmetric bilinear form $g:V \times V \rightarrow \mathbb{R}$. With this you can give a basis-independent mapping between $V$ and $V^*$. For any $\vec{v} \in V$ you can define a linear form $\underline{L}_{\vec{v}}$ by defining
$$\underline{L}_{\vec{v}}(\vec{w})=g(\vec{v},\vec{w}).$$
By definition of a non-degenerate bilinear form this (obviously linear) map $\vec{v} \mapsto \underline{L}_{\vec{v}}:V \rightarrow V^*$ is an isomorphism.

It's also easy to get from this "canonical definitions" ("canonical" meaning that we didn't use any basis to define the mapping) the usual Ricci calculus for the components. So let $\vec{b}_j$ and $\underline{b}^k$ be an arbitrary basis and its dual basis. Then the bilinear form is defined by its components wrt. the basis
$$g_{jk}=g(\vec{b}_j,\vec{b}_k)=g_{kj}.$$
Then you have
$$g(\vec{v},\vec{w})=g(v^j \vec{b}_j,w^k \vec{b}_k)=g_{jk} v^j w^k.$$
Then the above mapping is obviously given by
$$\underline{L}_{\vec{v}}(\vec{w})=L_{\vec{v},k} w^k=g_{jk} v^j w^k \; \Rightarrow \; L_{\vec{v},k}=g_{jk} v^j.$$
Finally you get
$$\underline{L}_{\vec{v}}=g_{jk} \underline{b}^k v^j:=v_k \underline{b}^k.$$
That means that
$$v_k=g_jk v^j.$$
Since the bilinear form is non-degenerate, the matrix $(g_{jk})$ is invertible and one defines the inverse matrix as $(g^{kl})$. It's uniquely defined by
$$g_{jk} g^{kl}=\delta_{j}^{l}, \quad g^{kl} g_{lj}=\delta_{j}^k.$$
Thus you have
$$v^j=g^{jk} v_k.$$
That's the usual rule for drawing indices up and down.

With these ideas you can work out all the manipulations needed for the Ricci calculus, including how the components of the various objects (tensors) transform. In the beginning this quite formal approach seems a bit cumbersome, but it helped at least me to understand the physicists' approach working only with components better. Particularly when in doubt, which transformation matrix you have to apply when transforming from one basis/dual basis pair to another, becomes very much more fail-safe when always writing everything in terms of invariant objects, i.e., the tensor components contracted with the corresponding basis/dual-basis vectors.

 

[pervect]

I suppose one could regard $x^{ab}$ as a dual space to $x_{ab}$, since it's a linear map from $x_{ab}$ to a scalar. But $x_{ab}$ is clearly a rank 2 tensor, not a rank 1 tensor, as it maps two vectors to a scalar, not one vector to a scalar. The key is the exact wording of the definition of the rank of a tensor, IMO. I'm not sure how it is or can be worded to avoid this confusion in general. In the context of general relativity, though, vectors are regarded as being 4 dimensional objects. $x_{ab}$, treated as a vector space, would have 16 dimensions as it has 16 components.

 

[vanhees71]

Of course tensors on of a fixed rank build a vector space as also do all tensors or all the alternating forms (completely antisymmetric tensors) etc.

 

[PeterDonis]

as also do all tensors

All tensors of all ranks, taken together? I don't think those form a vector space; how do you add tensors of different ranks?

all the alternating forms (completely antisymmetric tensors)

Same question here.

 

[Dale]

All tensors of all ranks, taken together? I don't think those form a vector space; how do you add tensors of different ranks?

You cannot even add tensors of the same rank representing physical quantities with different units. Like you cannot add the EM field tensor and the stress energy tensor even though they are the same rank.

I am sure he meant that each type of tensor forms its own vector space.

 

[vanhees71]

Good point. I was too quick and sloppy. Of course only tensors of the same rank form vector spaces.

 

[martinbn]

You can make all tensors into a vector space. Take the direct sum of all vector spaces of tensors of the same rank.

 

[PeterDonis]

You can make all tensors into a vector space. Take the direct sum of all vector spaces of tensors of the same rank.

Yes, this will work, but the elements of this vector space won't be tensors. They will be (infinite) tuples of tensors.