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Lecture 4 described vector fields on manifolds. We now look at tensors on smooth manifolds.
Suppose that v = (v_{1}, v_{2}, v_{3}) and w = (w_{1}, w_{2}, w_{3}) are vector fields on E_{3}. Then their tensor product is defined to consist of the nine quantities v_{i}w_{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}  =  x^{k} 
V^{k}  x^{m} 
W^{m}  =  x^{k} 
x^{m} 
V^{k} W^{m}  , 
and similarly,
^{i} _{j}  =  x^{k} 
^{j} 
V^{k} C_{m}  , 
and
_{i} _{j}  =  ^{j} 
^{j} 
C_{k} D_{m}  . 
We call these product fields "tensors" of type (2, 0), (1, 1), and (0, 2) respectively.
Definition 5.1 A tensor field of type (2, 0) on the ndimensional smooth manifold M associates with each chart x a collection of n^{2} smooth functions T^{ij}(x^{1}, x^{2}, . . . , x^{n}) which satisfy the transformation rules shown below. Similarly, we define tensor fields of type (0, 2), (1, 1), and, more generally, a tensor field of type (m, n). 
Some Tensor Transformation Rules

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
_{j}^{i}  = 

is, in fact a tensor field of type (1, 1). Indeed, one has
_{j}^{i}  =  x^{j} 
, 
and the latter quantities transform according to the rule
^{ij}  =  x^{k} 
x^{m} 
^{j} 

=  x^{k} 
^{j} 
^{k}_{m} 
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.  _{i}^{j} is independent of the chart used (the coordinates are the same as the barrd coordinates). Also, _{i}^{j} = _{j}^{i}. That is, it is a symmetric tensor.  
2. 

(c) We can make new tensor fields out of old ones by taking products of existing tensor fields in various ways. For example,
while
Specific examples of these involve the Kronecker delta, and are in the homework.
(d) If X is a contravariant vector field, then the functions X^{i}/x^{j}do not define a tensor. Indeed, let us check the transformation rule directly:
^{j}  = 


= 


= 

We will see more interesting examples later.
Proposition 5.3 (If It Looks Like a Tensor, It Is a Tensor)
Suppose that we are given smooth local functions g_{ij} with the property that for every pair of contravariant vector fields X^{i} and Y^{i}, the smooth functions g_{ij}X^{i}Y^{j} determine a scalar field. Then the g_{ij} determine a smooth tensor field of type (0, 2). 
Proof Since the g_{ij}X^{i}Y^{j} form a scalar field, we must have
On the other hand,
_{ij}^{i}^{j}  =  _{ij}X^{h}Y^{k}  x^{h} 
x^{k} 
by the transformation rules for contravariant vectors. Equating the righthand sides gives
g_{hk}X^{h}Y^{k}  =  _{ij}  x^{h} 
x^{k} 
X^{h}Y^{k }  .....................  (I) 
Now, if we could only cancel the terms X^{h}Y^{k}! Well, choose a point m M. It suffices to show that
g_{hk}  =  _{ij}  x^{h} 
x^{k} 
, 
when evaluated at the coordinates of m. However, by Example 4.3(c), we can arrange for vector fields X and Y such that
X^{i}(coordinates of m)  = 

, 
and
Y^{i}(coordinates of m)  = 

. 
Substituting these into equation (I) now gives the required transformation rule.
Example 5.4 Metric Tensor
Define a set of quantities g_{ij} by
g_{ij}  =  x^{i} 
.  x^{j} 
Question Why should I believe that this is a tensor?
Answer Let us invoke the above proposition: If X^{i} and Y^{j} are any contravariant fields on M, then their dot product X.Y is a scalar, and
X . Y  =  X^{i}  x^{i} 
.  Y^{j}  x^{j} 
=  g_{ij}X^{i}Y^{j}  . 
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. Suband superscripted quantities (other than coordinates) are understood to be tensors.
(a)  dt 
(b)  x^{j} 
(c)  x^{j} 
(d)  x^{i}x^{j} 
(e) 
2. (Rund, p. 95 #3.4) Show that if A_{j} is a type (0, 1) tensor, then
x^{h} 
  x^{k} 
is a type (0, 2) tensor.
3. Show that, if M and N are tensors of type (1, 1), then:
4. Let X be a contravariant vector field, and suppose that M is such that all changeofcoordinate maps have the form ^{i} = a^{ij}x^{j} + k^{i} for certain constants a^{ij} and k^{j}. (We call such a manifold affine.) Show that the functions X^{i}/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 skewsymmetric on the last pair of indices vanishes.
6. (Rund, p. 96, 3.16) If A_{kj} is a skewsymmetric tensor of type (0, 2), show that the quantities B_{rst} defined by
B_{rst}  =  x^{r} 
+  x^{s} 
+  x^{t} 
(a) are the components of a tensor; and
(b) are skewsymmetric 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
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.]
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