Let us now consider the *specific* nature of tensors on
four-dimensional space-time. Tensors of rank
^{18.1} are categorized (for each coordinate index) by their
transformation properties relative to a transformation of the underlying
coordinate system
as defined above. This transformation is
implicit in all the discussion below.

A **scalar** (tensor of rank zero) is unchanged by such a
transformation. *This is not a trivial statement!* It is trivial
for scalar numbers like
, no doubt, but in physics the *interesting* part of this requirement occurs when discussing the scalars
that result *algebraically* from fully contracting products of
tensors over all of their indices using the *metric tensor*. This
will be made quite clear below.

For a **vector** (tensor of rank one) we have two possibilities.
Either it transforms like the coordinate itself and we have a

**contravariant vector**- such that A^&alpha#alpha; = ^&alpha#alpha;x^&beta#beta; A^&beta#beta;

Alternatively, we have a

**covariant vector**- such that B_&alpha#alpha; = x^&beta#beta;^&alpha#alpha; B_&beta#beta;

In a moment we will see explicitly what exactly the difference is between these two types of first rank tensors. First, however, we should note that

**contravariant tensors of rank 2**- transform like F^&alpha#alpha;&beta#beta; = ^&alpha#alpha;x^&gamma#gamma; ^&beta#beta;x^&delta#delta; F^&gamma#gamma;&delta#delta; .

**covariant tensors of rank 2**- G_&alpha#alpha;&beta#beta; = x^&gamma#gamma;^&alpha#alpha; x^&delta#delta;^&beta#beta; G_&gamma#gamma;&delta#delta;

**mixed tensors of rank 2**- H^&alpha#alpha;_&beta#beta;= ^&alpha#alpha;x^&gamma#gamma; x^&delta#delta;^&beta#beta; H^&gamma#gamma;_&delta#delta;.

It is clearly a trivial exercise to determine the co/contra variant
transformation properties of higher rank tensors. We can form higher
rank tensors by means of an *outer* (dyadic) *product*, where we
simply take two tensors of some rank and multiply them out
componentwise, preserving products of any underlying basis vectors as
they occur. For example we can construct a second rank tensor by:
F^&alpha#alpha;&beta#beta; = A^&alpha#alpha;B^&beta#beta;
where
and
run over the full range of index values.
Note well that this defines a *square matrix* in this case of basis
vector dyads as objects such as
,
, ... occur.

One important question is whether *all* e.g. second rank tensors can
be written as products of first rank tensors. It is not the general
case that this is possible, but in many of our uses of these ideas in
physics it will be. In this case the generalized product forms a *division algebra* where we can *factor* e.g. second rank tensors
into first rank tensors in various ways. Division algebras are
discussed in the Mathematical Physics section as well, and interested
students should return there to read about *geometric algebras*, the
result of fully generalizing the notion of complex numbers to complex
spaces of arbitrary dimension while preserving the factorizability of
the algebraic objects.

In addition to *extending* the rank of tensor objects by forming
dyadic, triadic, or n-adic products of tensors, we can *reduce* the
rank of tensors by means of a process called *contraction*. A
contraction of two tensors is the result of setting two of the indices
(typically they must be a covariant/contravariant pair) to be equal and
performing the Einstein summation over the shared range. This reduces
the rank of the expression by one relative to that of its constituents,
hence the term ``contraction''. An expression can be contracted over
several components at a time when doing algebra so second rank tensors
can be contracted to form a 4-scalar, for example, or third rank tensors
can be contracted to first.

Our familiar notion of multiplying a vector by a matrix to produce a vector in proper tensor language is to form the outer product of the matrix (second rank tensor) and the vector (first rank tensor), set the rightmost indices to be equal and sum over that index to produce the resulting first rank tensor.

Hence we define our **scalar product** to be the **contraction** of
a covariant and contravariant vector.
B ·A = B_&alpha#alpha;A^&alpha#alpha;
Note that I've introduced a sort of ``sloppy'' convention that a single
quantity like
or
can be a four-vector in context. Clearly the
expression on the right side is less ambiguous!

Then:
B' ·A' & = & x^&gamma#gamma;^&alpha#alpha; B_&gamma#gamma;
^&alpha#alpha;x^&delta#delta; A^&delta#delta;

& = & x^&gamma#gamma;x^&delta#delta; B_&gamma#gamma;A^&delta#delta;

& = & &delta#delta;_&gamma#gamma;&delta#delta; B_&gamma#gamma;A^&delta#delta;

& = & B_&delta#delta;A^&delta#delta;= B ·A
and the desired invariance property is proved. Hmmm, that was pretty
easy! Maybe there is something to this notation thing after all!