Lecture 4 described vector fields on manifolds. We now look at tensors on smooth manifolds.

Suppose that v = (v₁, v₂, v₃) and w = (w₁, w₂, w₃) are vector fields on E₃. Then their tensor product is defined to consist of the nine quantities v_iw_j.

Let us see how such things transform. Thus, let V and W be contravariant, and let C and D be covariant. Then:

i j

=

i xk

Vk

j xm

Wm

=

i xk

j xm

Vk Wm

,

and similarly,

i j

=

i xk

xm j

Vk Cm

,

and

i j

=

xk j

xm j

Ck Dm

.

We call these product fields "tensors" of type (2, 0), (1, 1), and (0, 2) respectively.

Note A tensor field of type (1, 0) is just a contravariant vector field, while a tensor field of type (0, 1) is a covariant vector field. Similarly, a tensor field of type (0, 0) is a scalar field. Type (1, 1) tensors correspond to linear transformations in linear algebra.

Examples 5.2

(a) Of course, by definition, we can take tensor products of vector fields to obtain tensor fields, as we did above in Definition 4.1.

(b) The Kronecker Delta Tensor, given by

ji

=

1   if j = i 0 if j i

is, in fact a tensor field of type (1, 1). Indeed, one has

ji

=

xi xj

,

and the latter quantities transform according to the rule

	i j	=	i xk	xk xm	xm j
		=	i xk	xm j	km

whence they constitute a tensor field of type (1, 1).

Question OK, so is this how it works: Given a point p of the manifold and a chart x at p this strange object assigns the n^2< quantities _i^j; that is, the identity matrix, regardless of the chart we chose?
Answer Yes.

Question But how can we interpret this strange object?
Answer Just as a covariant vector field converts contravariant fields into scalars (see Lecture 3) we shall see that a type (1,1) tensor converts contravariant fields to other contravariant fields. This particular tensor does nothing: put in a specific vector field V, out comes the same vector field. In other words, it is the identity transformation.

Notes

1.	ij is independent of the chart used (the coordinates are the same as the barrd coordinates). Also, ij = ji. That is, it is a symmetric tensor.
2.	i xj xj k = i k = ik

(c) We can make new tensor fields out of old ones by taking products of existing tensor fields in various ways. For example,

Mⁱ_jk N^pq_rs is a tensor of type (3, 4),

while

Mⁱ_jk N^jk_rs is a tensor of type (1, 2).

Specific examples of these involve the Kronecker delta, and are in the homework.

(d) If X is a contravariant vector field, then the functions Xⁱ/x^jdo not define a tensor. Indeed, let us check the transformation rule directly:

i j	=		j Xk i xk
	=		xh Xk i xk xh j
	=		Xk xh i xk xh j + Xk 2i xhxk

We will see more interesting examples later.

Proof Since the g_ijXⁱY^j form a scalar field, we must have

_ij^ij = g_hkX^hY^k.

On the other hand,

ijij

=

ijXhYk

i xh

j xk

by the transformation rules for contravariant vectors. Equating the right-hand sides gives

ghkXhYk

=

ij

i xh

j xk

XhYk

.....................

(I)

Now, if we could only cancel the terms X^hY^k! Well, choose a point m M. It suffices to show that

ghk

=

ij

i xh

j xk

,

when evaluated at the coordinates of m. However, by Example 4.3(c), we can arrange for vector fields X and Y such that

Xi(coordinates of m)

=

1   if i = h 0 if i h

,

and

Yi(coordinates of m)

=

1   if i = k 0 if i k

.

Substituting these into equation (I) now gives the required transformation rule.

Example 5.4 Metric Tensor

Define a set of quantities g_ij by

gij

=

xi

.

xj

Question Why should I believe that this is a tensor?
Answer Let us invoke the above proposition: If Xⁱ and Y^j are any contravariant fields on M, then their dot product X.Y is a scalar, and

X . Y

=

Xi

xi

.

Yj

xj

=

gijXiYj

.

Thus, by Proposition 5.3, it is a type (0, 2) tensor. We call this tensor "the metric tensor inherited from the imbedding of M in E_s."

Exercise Set 5

1. Compute the transformation rules for each of the following, and hence decide whether or not they are tensors. Sub-and superscripted quantities (other than coordinates) are understood to be tensors.

(a)

dXij dt

(b)

xi xj

(c)

Xi xj

(d)

2 xixj

(e)

2xp xixj

2. (Rund, p. 95 #3.4) Show that if A_j is a type (0, 1) tensor, then

Ak xh

-

Ah xk

is a type (0, 2) tensor.

3. Show that, if M and N are tensors of type (1, 1), then:

(a) Mⁱ_jN^p_q is a tensor of type (2, 2)
(b) Mⁱ_jN^j_q is a tensor of type (1, 1)
(c) Mⁱ_jN^j_j is a tensor of type (0, 0) (that is, a scalar field)

4. Let X be a contravariant vector field, and suppose that M is such that all change-of-coordinate maps have the form ⁱ = a^ijx^j + kⁱ for certain constants a^ij and k^j. (We call such a manifold affine.) Show that the functions Xⁱ/x^j define a tensor field of type (1, 1).

5. (Rund, p. 96, 3.12) If B^ijk = -B^jki, show that B^ijk = 0. Deduce that any type (3, 0) tensor that is symmetric on the first pair of indices and skew-symmetric on the last pair of indices vanishes.

6. (Rund, p. 96, 3.16) If A_kj is a skew-symmetric tensor of type (0, 2), show that the quantities B_rst defined by

Brst

=

Ast xr

+

Atr xs

+

Ars xt

(a) are the components of a tensor; and
(b) are skew-symmetric in all pairs in indices.
(c) How many independent components does B_rst have?

8. Suppose that C^ij is a type (2, 0) tensor, and that, regarded as an nÀn matrix C, it happens to be invertible in every coordinate system. Define a new collection of functions, D_ij by taking

D_ij = C^-1_ij,

the ij the entry of C^-1 in every coordinate system. Show that D_ij, is a type (0, 2) tensor. [Hint: Write down the transformation equation for C^ij and invert everything in sight.]