2.3 Tensors

For any vector space V , the dual vector space $V^{\ast }$ is the space of all linear maps from V to R.

This is a standard mathematical construction, but even if you haven’t seen it before it should resonate with something you know from quantum mechanics. There we have states in a Hilbert space with kets $|\psi ⟩\in \mathcal{H}$ and a dual Hilbert space with bras $|\varphi |\in {\mathcal{H}}^{\ast }$ Any bra can be viewed as a map $\varphi |:\mathcal{H}\to \mathbf{R}$ defined by $⟨\varphi |(|\psi ⟩)=⟨\varphi |\psi ⟩$

In general, suppose that we are given a basis $\{e_{\mu },\mu =1,\dots ,n\}$ of V. Then we can introduce a dual basis $\{f^{\mu },\mu =1,\dots ,n\}$ for $V^{\ast }$ defined by

$$f^{\nu }(e_{\mu })={\delta }_{\mu }^{\nu }$$

A general vector in V can be written as $X=X^{\mu }{e}_{\mu }$ and $f^{\nu }(X)=X^{\mu }{f}^{\nu }(e_{\mu })=X^{\nu }$ Given a basis, this construction provides an isomorphism between V and $V^{\ast }$ given by $e_{\mu }\to f^{\mu }$ . But the isomorphism is basis dependent. Pick a different basis, and you’ll get a different map.

We can repeat the construction and consider $(V^{\ast }{)}^{\ast }$ , which is the space of all linear maps from $V^{\ast }$ to R. But this space is naturally isomorphic to $V$ , meaning that the isomorphism is independent of the choice of basis. To see this, suppose that $X\in V$ and $\omega \in V^{\ast }$ . This means that $\omega (X)\in \mathbf{R}$ . But we can equally as well view $X\in (V^{\ast }{)}^{\ast }$ and define $X(\omega )=\omega (X)\in \mathbf{R}$ . In this sense, $(V^{\ast }{)}^{\ast }=V$

2.3.1 Covectors and One-Forms

At each point $p\in M$ , we have a vector space $T_p(M)$ . The dual of this space, $T_p^{\ast }(M)$ is called the cotangent space at $p,$ and an element of this space is called a cotangent vector, sometimes shortened to covector. Given a basis $\{e_{\mu }\}$ of $T_p(M)$ , we can introduce the dual basis $\{f^{\mu }\}$ for $T_p^{\ast }(M)$ and expand any co-vector as $\omega ={\omega }_{\mu }{f}^{\mu }$

We can also construct fields of cotangent vectors, by picking a member of $T_p^{\ast }(M)$ for each point $p$ in a smooth manner. Such a cotangent field is better known as a one-form; they map vector fields to real numbers. The set of all one-forms on M is denoted ${\Lambda }^1(M)$

There is a particularly simple way to construct a one-form. Take a function $f\in$ $C^{\infty }(M)$ and define $df\in {\Lambda }^1(M)$ by

$$df(X)=X(f)\text{\hspace{1em}(2.16)}$$

We can use this method to build a basis for ${\Lambda }^1(M)$ . We introduce coordinates $x^{\mu }$ on M with the corresponding coordinate basis $e_{\mu }=\partial /\partial x^{\mu }$ of vector fields, which we often write in shorthand as $\partial /\partial x^{\mu }\equiv {\partial }_{\mu }$ . We then simply take the functions $f=x^{\mu }$ which, from (2.16), gives

$$d{x}^{\mu }({\partial }_{\nu })={\partial }_{\nu }(x^{\mu })={\delta }_{\nu }^{\mu }$$

This means that $f^{\mu }=d{x}^{\mu }$ provides a basis for ${\Lambda }^1(M)$ , dual to the coordinate basis $\partial /{\partial }_{\mu }$ . In general, an arbitrary one-form $\omega \in {\Lambda }^1(M)$ can then be expanded as

$$\omega ={\omega }_{\mu }d{x}^{\mu }$$

In such a basis the one-form df takes the form

$$df=\frac{\partial f}{\partial x^{\mu }}d{x}^{\mu }\text{\hspace{1em}(2.17)}$$

To see this, we simply need to evaluate

$$df(X)=\frac{\partial f}{\partial x^{\mu }}d{x}^{\mu }(X^{\nu }{\partial }_{\nu })=X^{\mu }\frac{\partial f}{\partial x^{\mu }}=X(f)$$

which agrees with the expected answer (2.16).

As with vector fields, we can look at what happens if we change coordinates. Given two different charts, $\varphi =(x^1,\dots ,x^n)$ and $\tilde{\varphi }=({\tilde{x}}^1,\dots ,{\tilde{x}}^n)$ , we know that the basis for vector fields changes as (2.5),

$$\frac{\partial }{\partial {\tilde{x}}^{\mu }}=\frac{\partial x^{\nu }}{\partial {\tilde{x}}^{\mu }}\frac{\partial }{\partial x^{\nu }}$$

We should take the basis of one-forms to transform in the inverse manner,

$$d{\tilde{x}}^{\mu }=\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\nu }}d{x}^{\nu }\text{\hspace{1em}(2.18)}$$

This then ensures that

$$d{\tilde{x}}^{\mu }\left(\frac{\partial }{\partial {\tilde{x}}^{\nu }}\right)=\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\rho }}d{x}^{\rho }\left(\frac{\partial x^{\sigma }}{\partial {\tilde{x}}^{\nu }}\frac{\partial }{\partial x^{\sigma }}\right)=\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\rho }}\frac{\partial x^{\sigma }}{\partial {\tilde{x}}^{\nu }}d{x}^{\rho }\left(\frac{\partial }{\partial x^{\sigma }}\right)=\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\rho }}\frac{\partial x^{\sigma }}{\partial {\tilde{x}}^{\nu }}{\delta }_{\sigma }^{\rho }$$

But this is just the multiplication of a matrix and its inverse,

$$\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\rho }}\frac{\partial x^{\rho }}{\partial {\tilde{x}}^{\nu }}={\delta }_{\nu }^{\mu }$$

So we find that

$$d{\tilde{x}}^{\mu }\left(\frac{\partial }{\partial {\tilde{x}}^{\nu }}\right)={\delta }_{\nu }^{\mu }$$

as it should. We can then expand a one-form $\omega$ in either of these two bases,

$$\omega ={\omega }_{\mu }d{x}^{\mu }={\tilde{\omega }}_{\mu }d{\tilde{x}}^{\mu }\mathrm{with}{\tilde{\omega }}_{\mu }=\frac{\partial x^{\nu }}{\partial {\tilde{x}}^{\mu }}{\omega }_{\nu }\text{\hspace{1em}(2.19)}$$

In the annoying terminology that I can never remember, components of vectors that transform this way are said to be covariant. Note that, as with vector fields, the placement of the indices means that (2.18) and (2.19) are pretty much the only things that you can write down that make sense.

2.3.2 The Lie Derivative Revisited

In Section 2.2.4, we explained how to construct the Lie derivative, which differentiates a vector field in the direction of a second vector field X. This same idea can be adapted to one-forms.

Under a map $\phi :M\to N$ , we saw that a vector field X on M can be pushed forwards to a vector field ${\phi }_{\ast }X$ on N. In contrast, one-forms go the other way: given a one-form $\omega$ on $N$ , we can pull this back to a one-form $\left({\phi }^{\ast }\omega \right)$ on $M.$ , defined by

$$({\phi }^{\ast }\omega )(X)=\omega ({\phi }_{\ast }X)$$

If we introduce coordinates $x^{\mu }$ on M and $y^{\alpha }$ on N then the components of the pull-back are given by

$$({\phi }^{\ast }\omega {)}_{\mu }={\omega }_{\alpha }\frac{\partial y^{\alpha }}{\partial x^{\mu }}\text{\hspace{1em}(2.20)}$$

We now define the Lie derivative ${\mathcal{L}}_X$ acting on one-forms. Again, we use $X$ to generate a flow ${\sigma }_t:M\to M$ which, using the pull-back, allows us to compare one-forms at different points. We will denote the cotangent vector $\omega (p)$ as ${\omega }_p$ . The Lie derivative of a one-form ω is then defined as

$${\mathcal{L}}_X\omega =\underset{t\to 0}{\lim }\frac{({\sigma }_t^{\ast }\omega {)}_p-{\omega }_p}{t}\text{\hspace{1em}(2.21)}$$

Note that we pull-back with the map ${\sigma }_t$ . This is to be contrasted with (2.13) where we pushed forward the tangent vector with the map ${\sigma }_{-t}$ and, as we now show, this difference in minus sign manifests itself in the expression for the Lie derivative. The infinitesimal map ${\sigma }_t$ acts on coordinates as $x^{\mu }(t)=x^{\mu }(0)+t{X}^{\mu }+\text{ . . . s o }$ , from (2.20), the pull-back of a basis vector $d{x}^{\mu }$ is

$${\sigma }_t^{\ast }d{x}^{\mu }=\left({\delta }_{\nu }^{\mu }+t\frac{\partial X^{\mu }}{\partial x^{\nu }}+\dots \right)d{x}^{\nu }$$

Acting on the coordinate basis, we then have

$${\mathcal{L}}_X(d{x}^{\mu })=\frac{\partial X^{\mu }}{\partial x^{\nu }}d{x}^{\nu }$$

which indeed differs by a minus sign from the corresponding result (2.14) for tangent vectors. Acting on a general one-form $\omega ={\omega }_{\mu }d{x}^{\mu }$ , the Lie derivative is

$$\begin{array}{c}{\mathcal{L}}_X\omega =({\mathcal{L}}_X{\omega }_{\mu })d{x}^{\mu }+{\omega }_{\nu }{\mathcal{L}}_X(d{x}^{\nu })\\ =(X^{\nu }{\partial }_{\nu }{\omega }_{\mu }+{\omega }_{\nu }{\partial }_{\mu }{X}^{\nu })d{x}^{\mu }\end{array}\text{\hspace{1em}(2.22)}$$

We’ll return to discuss one-forms (and other forms) more in Section 2.4.

2.3.3 Tensors and Tensor Fields

A tensor of rank $(r,s)$ at a point $p\in M$ is defined to be a multi-linear map

$$T:\stackrel{r}{\overbrace{T_p^{\ast }(M)\times \dots \times T_p^{\ast }(M)}}\times \stackrel{s}{\overbrace{T_p(M)\times \dots \times T_p(M)}}\to \mathbf{R}$$

Such a tensor is said to have total rank $r+s.$

We’ve seen some examples already. A cotangent vector in $T_p^{\ast }(M)$ is a tensor of type $(0,1)$ , while a tangent vector in $T_p(M)$ is a tensor of type (1, 0) (using the fact that $T_p(M)=T_p^{\ast \ast }(M))$ ).

As before, we define a tensor field to be a smooth assignment of an $(r,s)$ tensor to every point $p\in M$

Given a basis $\{e_{\mu }\}$ for vector fields and a dual basis $\{f^{\mu }\}$ for one-forms, the components of the tensor are defined to be

$$T_{{\nu }_1\dots {\nu }_s}^{{\mu }_1\dots {\mu }_r}=T(f^{{\mu }_1},\dots ,f^{{\mu }_r},e_{{\nu }_1},\dots ,e_{{\nu }_s})$$

Note that we deliberately write the string of lower indices after the upper indices. In some sense this is unnecessary, and we don’t lose any information by writing $T_{{\nu }_1...{\nu }_s}^{{\mu }_1...{\mu }_r}$ Nonetheless, we’ll see later that it’s a useful habit to get into.

On a manifold of dimension n, there are $n^{r+s}$ such components. For a tensor field, each of these is a function over M.

As an example, consider a rank $(2,1)$ tensor. This takes two one-forms, say ω and $\eta ,$ together with a vector field X, and spits out a real number. In a given basis, this number is

$$T(\omega ,\eta ,X)=T({\omega }_{\mu }{f}^{\mu },{\eta }_{\nu }{f}^{\nu },X^{\rho }{e}_{\rho })={\omega }_{\mu }{\eta }_{\nu }{X}^{\rho }T(f^{\mu },f^{\nu },e_{\rho })=T_{\rho }^{\mu \nu }{\omega }_{\mu }{\eta }_{\nu }{X}^{\rho }$$

Every manifold comes equipped with a natural $(1,1)$ tensor called δ. This takes a one-form $\omega$ and a vector field X and spits out the real number

$$\delta (\omega ,X)=\omega (X)\Rightarrow \delta (f^{\mu },e_{\nu })=f^{\mu }(e_{\nu })={\delta }_{\nu }^{\mu }$$

which is simply the Kronecker delta.

As with vector fields and one-forms, we can ask how the components of a tensor transform. We will work more generally than before. Consider two bases for the vector fields, $\{e_{\mu }\}$ and $\{{\tilde{e}}_{\mu }\}$ , not necessarily coordinate bases, related by

$${\tilde{e}}_{\nu }=A_{\nu }^{\mu }{e}_{\mu }$$

for some invertible matrix A. The respective dual bases are $\{f^{\mu }\}$ and $\left\{{\tilde{f}}^{\mu }\right\}$ are then related by

$${\tilde{f}}^{\rho }=B_{\sigma }^{\rho }{f}^{\sigma }$$

such that

$${\tilde{f}}^{\rho }({\tilde{e}}_{\nu })=A_{\nu }^{\mu }{B}_{\sigma }^{\rho }{f}^{\sigma }(e_{\mu })=A_{\nu }^{\mu }{B}_{\mu }^{\rho }={\delta }_{\nu }^{\rho }\Rightarrow B_{\mu }^{\rho }=(A^{-1}{)}_{\mu }^{\rho }$$

The lower components of a tensor then transform by multiplying by A, and the upper components by multiplying by $B=A^{-1}$ . So, for example, a rank (1, 2) tensor transforms as

$${\tilde{T}}_{\rho \nu }^{\mu }=B_{\sigma }^{\mu }{A}_{\rho }^{\tau }{A}_{\nu }^{\lambda }{T}_{\tau \lambda }^{\sigma }\text{\hspace{1em}(2.23)}$$

When we change between coordinate bases, we have

$$A_{\nu }^{\mu }=\frac{\partial x^{\mu }}{\partial {\tilde{x}}^{\nu }}\mathrm{and}{B}_{\nu }^{\mu }=(A^{-1}{)}_{\nu }^{\mu }=\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\nu }}$$

You can check that this coincides with our previous results (2.6) and (2.19).

Operations on Tensor Fields

There are a number of operations that we can do on tensor fields to generate further tensors.

First, we can add and subtract tensors fields, or multiply them by functions. This is the statement that the set of tensors at a point $p\in M$ forms a vector space.

Next, there is a way to multiply tensors together to give a tensor of a different type. Given a tensor S of rank $(p,q)$ and a tensor T of rank $(r,s)$ , we can form the tensor product, $S\otimes T$ which a new tensor of rank $(p+r,q+s)$ , defined by

$$\begin{array}{c}S\otimes T({\omega }_1,\dots ,{\omega }_p,{\eta }_1,\dots ,{\eta }_r,X_1,\dots X_q,Y_1,\dots ,Y_s)\\ =S({\omega }_1,\dots ,{\omega }_p,X_1,\dots X_q)T({\eta }_1,\dots ,{\eta }_r,Y_1,\dots ,Y_s)\end{array}$$

In terms of components, this reads

$$(S\otimes T{)}^{\begin{array}{c}{\mu }_1\dots {\mu }_p{\nu }_1\dots {\nu }_r\\ {\rho }_1\dots {\rho }_q{\sigma }_1\dots {\sigma }_s\end{array}}=S_{{\rho }_1\dots {\rho }_q}^{{\mu }_1\dots {\mu }_p}{T}_{{\sigma }_1\dots {\sigma }_s}^{{\nu }_1\dots {\nu }_r}\text{\hspace{1em}(2.24)}$$

Given an $(r,s)$ tensor $T_{\cdot }$ , we can also construct a tensor of lower rank $(r-1,s-1)$ by contraction. To do this, simply replace one of $T_p^{\ast }(M)$ entries with a basis vector $f^{\mu }$ , and the corresponding $T_p(M)$ entry with the dual vector $e_{\mu }$ and then sum over $\mu =1,\dots ,n$ . So, for example, given a rank (2, 1) tensor T we can construct a rank (1, 0) tensor S by

$$S(\omega )=T(\omega ,f^{\mu },e_{\mu })$$

Alternatively, we could construct a (typically) different (1, 0) tensor by contracting the other argument, $S^{\prime }(\omega )=T(f^{\mu },\omega ,e_{\mu })$ . Written in terms of components, contraction simply means that we put an upper index equal to a lower index and sum over them,

$$S^{\mu }=T_{\nu }^{\mu \nu }\mathrm{and}{S}^{\prime \mu }=T_{\nu }^{\nu \mu }$$

Our next operation is symmetrisation and anti-symmetrisation. For example, given a (0, 2) tensor $T$ we decompose it into two (0, 2) tensors, in which the arguments are either symmetrised or anti-symmetrised,

$$\begin{array}{rl}& S(X,Y)=\frac{1}{2}(T(X,Y)+T(Y,X))\\ & A(X,Y)=\frac{1}{2}(T(X,Y)-T(Y,X))\end{array}$$

In index notation, this becomes

$$S_{\mu \nu }=\frac{1}{2}(T_{\mu \nu }+T_{\nu \mu })\mathrm{and}{A}_{\mu \nu }=\frac{1}{2}(T_{\mu \nu }-T_{\nu \mu })$$

which is just like taking the symmetric and anti-symmetric part of a matrix. We will work with these operations frequently enough to justify introducing some new notation. We define

$$T_{(\mu \nu )}=\frac{1}{2}(T_{\mu \nu }+T_{\nu \mu })\mathrm{and}{T}_{[\mu \nu ]}=\frac{1}{2}(T_{\mu \nu }-T_{\nu \mu })$$

These operations generalise to other tensors. For example,

$$T_{\sigma }^{(\mu \nu )\rho }=\frac{1}{2}(T_{\sigma }^{\mu \nu \rho }+T_{\sigma }^{\nu \mu \rho })$$

We can also symmetrise or anti-symmetrise over multiple indices, provided that these indices are either all up or all down. If we (anti)-symmetrise over $p$ objects, then we divide by $p!.$ , which is the number of possible permutations. This normalisation ensures that if we start with a tensor which is already, say, symmetric then further symmetrising doesn’t affect it. In the case of anti-symmetry, we weight each term with the sign of the permutation. So , for example,

$$T_{(\nu \rho \sigma )}^{\mu }=\frac{1}{3!}(T_{\nu \rho \sigma }^{\mu }+T_{\rho \nu \sigma }^{\mu }+T_{\rho \sigma \nu }^{\mu }+T_{\sigma \rho \nu }^{\mu }+T_{\sigma \nu \rho }^{\mu }+T_{\nu \sigma \rho }^{\mu })$$

and

$$T_{[\nu \rho \sigma ]}^{\mu }=\frac{1}{3!}(T_{\nu \rho \sigma }^{\mu }-T_{\rho \nu \sigma }^{\mu }+T_{\rho \sigma \nu }^{\mu }-T_{\sigma \rho \nu }^{\mu }+T_{\sigma \nu \rho }^{\mu }-T_{\nu \sigma \rho }^{\mu })$$

There will be times when, annoyingly, we will wish to symmetrise (or anti-symmetrise) over indices which are not adjacent. We introduce vertical bars to exclude certain indices from the symmetry procedure. So, for example,

$$T_{[\nu |\rho |\sigma ]}^{\mu }=\frac{1}{2}(T_{\nu \rho \sigma }^{\mu }-T_{\sigma \rho \nu }^{\mu })$$

Finally, given a smooth tensor field $T$ of any rank, we can always take the Lie derivative with respect to a vector field X. As we’ve seen previously, under a map $\phi :M\to N$ 2 vector fields are pushed forwards and one-forms are pulled-back. In general, this leaves a tensor of mixed type unsure where to go. However, if $\phi$ is a diffeomorphism then we also have ${\phi }^{-1}:N\to M$ and this allows us to define the push-forward of a tensor $T$ from M to N. This acts on one-forms $\omega \in {\Lambda }^1(N)$ and vector fields $X\in \mathfrak{X}(N)$ and is given by

$$({\phi }_{\ast }T)({\omega }_1,\dots ,{\omega }_r,X_1,\dots ,X_s)=T({\phi }^{\ast }{\omega }_1,\dots ,{\phi }^{\ast }{\omega }_r,({\phi }_{\ast }^{-1}{X}_1),\dots ,({\phi }_{\ast }^{-1}{X}_s))$$

Here ${\phi }^{\ast }\omega$ are the pull-backs of $\omega$ from N to $M$ , while ${\phi }_{\ast }^{-1}X$ are the push-forwards of X from N to M.

The Lie derivative of a tensor T along X is then defined as

$${\mathcal{L}}_{X}T=\underset{t\to 0}{\lim }\frac{(({\sigma }_{-t}{)}_{\ast }T{)}_p-T_p}{t}$$

where ${\sigma }_t$ is the flow generated by X. This coincides with our earlier definitions for vector fields in (2.13) and for one-forms in (2.21). (The difference in the ${\sigma }_{-t}$ vs ${\sigma }_t$ minus sign in (2.13) and (2.21) is now hiding in the inverse push-forward ${\phi }_{\ast }^{-1}$ that appears in the definition ${\phi }_{\ast }T.)$

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正文：英文 · OCR 机器稿 · 待校对

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