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.