Let be an
-dimensional Euclidean space. In Lecture 22, we constructed the Euclidean spaces
of degree
tensors, as well as the subspaces
and
of
consisting of symmetric and antisymmetric tensors, respectively. Let us recall the formulas for the scalar product in these vector spaces induced by the scalar product
in
:
To be precise, the first of these formulas is a definition, while the second and third formulas are consequences of this definition and the fact that and
are subspaces of
In this lecture we focus on the Euclidean space which turns out to be the most important of the three. In fact, we will now be dealing with the antisymmetric tensor powers of
as standalone objects, rather than viewing them as subspaces of the tensor powers
and as such it will be more convenient to renormalize the scalar product to
dropping the factor of which was previously hanging around outside the sum. We are free to do this, since rescaling a scalar product by any positive constant yields a scalar product.
We have already seen that is a linearly dependent set in
if and only if
where
denotes the zero tensor of degree Equivalently,
is a linearly dependent set in
if and only if
Today, we will expand on this by showing that length in provides a good definition of volume in
Definition 1: For any vectors the corresponding parallelepiped is the subset of
defined by
We define the volume of this parallelepiped by
According to Definition 1, we should interpret the length of the tensor to be the volume of the parallelepiped spanned by its factors. This makes sense, in that the length of
is zero if and only if
is a linearly dependent set in
which means that
lies in a subspace of
whose dimension strictly less than
. This corresponds to our intuition that the length of a point is zero, the area of a line segment is zero, and the volume of a parallelogram is zero. Indeed, Theorem 2 from Lecture 22 applied to Definition 1 yields the following statement.
Theorem 1: The set is a basis in
if and only if
As an example, let us consider the case i.e.
is a
-dimensional vector space. Let
We then have that
consists of all vectors of the form
with
which can be visualized as the line segment joining
to
in
The volume of
is then
which is the length of this line segment. So, according to our definition, volume and length are the same thing in one dimension, which seems reasonable.
As another example, let us consider the case i.e.
is a
-dimensional Euclidean space. Let
Then,
can be visualized as the set of all vectors in
which lie on or inside the parallelogram with vertices
According to our definition, the volume of is the number
Let us evaluate this number more explicitly. Let be an orthonormal basis of
We then have
so that
Since
we conclude that
There are a couple of things we can notice observe about this number. First, we can also write it as
as we will see momentarily, this is because is an orthonormal basis of
Second, if you are familiar with the definition of the determinant of a
matrix,
it is apparent that is equal to the absolute value of the determinant of the matrix
whose columns are the coordinates of the vectors relative to the basis
This coincidence will also be explained shortly. First, however, we want to ask a natural question: what if we have done the above computation using a different orthonormal basis
of
? The same calculations would then have led us to the formula
which is apparently different from out first formula. But volume should be a geometric entity, and its computation should not depend on the choice of coordinates, i.e. on a choice of basis. We can see our way through this by considering the orthogonal transformation uniquely determined by
We then see that
which is equivalent to the assertion that
This makes perfect geometric sense: since orthogonal transformations preserve lengths and angles, the parallelogram is just a rotated and/or flipped copy of
which naturally has the same area.
The key step in understanding the linear algebra underlying our definition of -dimensional volume is finding an orthonormal basis of
the space of antisymmetric tensors of degree
We have solved this problem for
the space of all tensors of degree
back in Lecture 22, we showed that any orthonormal basis
in
induces a corresponding orthonormal basis
in so that
To obtain the analogous result for
we need to introduce the subset of
consisting of increasing functions:
Theorem 2: For any the set
is an orthonormal basis in For all
the only antisymmetric tensor of degree
is the zero tensor, i.e.
Proof: First we prove that spans
To do this, it is sufficient to show that any simple tensor
is a linear combination of the tensors in But this is clear from the multilinearity of the wedge product: we have
Now, since a wedge product which contains two copies of the same vector is zero, the only nonzero terms in the above sum are those corresponding to injective functions, i.e. those such that the numbers
are pairwise distinct. If
there are no such functions, and consequently
latex d \leq n,$ then for any injection
the factors of the nonzero tensor
can be sorted so that the distinct indices are in increasing order; doing this will scale the original tensor by a factor of
Now we prove that is an orthonormal set in
(in particular, this will imply linear independence). Let
be distinct increasing functions. We then have that
Since there is no permutation which can transform the list
of distinct numbers into the different list
of distinct numbers, every term in the sum is zero. This proves that
is an orthogonal set in
To prove that each element of
is of unit length, repeat the above argument with
Then, the only nonzero term in the sum is that corresponding to the identity permutation:
— Q.E.D.
Corollary 1: The dimension of is
Proof: Any increasing function corresponds to a unique cardinality
subset
of
. By definition, the binomial coefficient
is the number of such sets.
— Q.E.D.
Although it may not seem so, the most important consequence of Corollary 1 is that
Indeed, according to Theorem 1, the set consisting of the single tensor
is a basis in This means that every antisymmetric tensor of degree
is a scalar multiple of
so that in particular for any vectors
we have
for some scalar the value of which depends on the vectors
Indeed, since
is an orthonormal basis in
we have
We thus have the following alternative description of volume.
Proposition 1: For any orthonormal basis in
we have
Proof: We will give two proofs, one computational and one conceptual.
Computational:
Conceptual: by Cauchy-Schwarz,
We know that equality holds in the Cauchy-Schwarz inequality precisely when linear dependence holds, which is the case here because Thus
— Q.E.D.
Any tensor of the form
where is an orthonormal basis in
, is called a unit volume tensor for
Indeed, for such a tensor we have
corresponding exactly to the fact that
is an
-dimensional unit box. In fact, because
is
-dimensional, there are only two unit volume tensors: if
satisfies
then the only other tensor in
with this property is
Definition 3: A triple consisting of a finite-dimensional vector space
a scalar product
on
, and a unit volume tensor
is called an oriented Euclidean space.
For example, consider the case of the Euclidean plane, i.e. with the standard scalar product. Giving the Euclidean plane an orientation means choosing one of the two tensors
Choosing the first tensor gives the Euclidean plane a counterclockwise orientation, while choosing the second gives the plane a clockwise orientation.
Definition 2: Given a unit volume tensor the corresponding determinant is the function
defined by
Thus is the oriented volume of the parallelepiped
meaning that it coincides with
up to a factor of
So, “oriented volume function” might be a more natural name than “determinant.” However, the term determinant is also appropriate, given the following result (which is equivalent to Theorem 1 above).
Theorem 3: The set is a basis in
if and only if
It is important to realize that the Euclidean space actually supports two distinct determinant functions, one for each of the two unit volume tensors in the one-dimensional space
For example, if
and we have two orthonormal bases
and
then we have two corresponding determinant functions
which may or may not coincide. For example, if
then we have
In Lecture 24, we shall discuss determinants of operators on Euclidean space, which are closely related to — but not quite the same as — oriented volumes of parallelepipeds in Euclidean space.
There is a spot under the proof of THM 2 where the latex did not quite work. It is right after where it explained V1^..^Vd=0 if d is greater than n.
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