# Math 31AH: Lecture 23

Let $\mathbf{V}$ be an $n$-dimensional Euclidean space. In Lecture 22, we constructed the Euclidean spaces $\mathbf{V}^{\otimes d}$ of degree $d$ tensors, as well as the subspaces $\mathbf{V}^{\vee d}$ and $\mathbf{V}^{\wedge d}$ of $\mathbf{V}^{\otimes d}$ 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 $\langle \cdot,\cdot \rangle$ in $\mathbf{V}$:

$\langle \mathbf{v}_1 \otimes \dots \otimes \mathbf{v}_d,\mathbf{w}_1 \otimes \dots \otimes \mathbf{w}_d \rangle = \prod\limits_{i=1}^d \langle \mathbf{v}_i,\mathbf{w}_i \rangle \\ \langle \mathbf{v}_1 \vee \dots \vee \mathbf{v}_d,\mathbf{w}_1 \vee \dots \vee \mathbf{w}_d \rangle = \frac{1}{d!}\sum\limits_{\pi \in \mathrm{S}(d)}\prod\limits_{i=1}^d \langle \mathbf{v}_i,\mathbf{w}_{\pi(i)} \rangle \\ \langle \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d,\mathbf{w}_1 \wedge \dots \wedge \mathbf{w}_d \rangle = \frac{1}{d!}\sum\limits_{\pi \in \mathrm{S}(d)} \mathrm{sgn}(\pi)\prod\limits_{i=1}^d \langle \mathbf{v}_i,\mathbf{w}_{\pi(i)} \rangle.$

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 $\mathbf{V}^{\vee d}$ and $\mathbf{V}^{\wedge d}$ are subspaces of $\mathbf{V}^{\otimes d}.$

In this lecture we focus on the Euclidean space $\mathbf{V}^{\wedge d},$ which turns out to be the most important of the three. In fact, we will now be dealing with the antisymmetric tensor powers of $\mathbf{V}^{\wedge d}$ as standalone objects, rather than viewing them as subspaces of the tensor powers $\mathbf{V}^{\otimes d},$ and as such it will be more convenient to renormalize the scalar product to

$\langle \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d,\mathbf{w}_1 \wedge \dots \wedge \mathbf{w}_d \rangle = \sum\limits_{\pi \in \mathrm{S}(d)} \mathrm{sgn}(\pi)\prod\limits_{i=1}^d \langle \mathbf{v}_i,\mathbf{w}_{\pi(i)} \rangle,$

dropping the factor of $\frac{1}{d!}$ 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 $\{ \mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly dependent set in $\mathbf{V}$ if and only if

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d = \mathbf{0},$

where

$\mathbf{0}=\mathbf{0}_{\mathbf{V}^{\otimes d}} = \underbrace{\mathbf{0}_\mathbf{V} \otimes \dots \otimes \mathbf{0}_\mathbf{V}}_{d \text{ factors}}$

denotes the zero tensor of degree $d.$ Equivalently, $\{\mathbf{v}_1,\dots,\mathbf{v}_d\}$ is a linearly dependent set in $\mathbf{V}$ if and only if

$\|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d\| = \sqrt{\langle \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d \rangle}=0.$

Today, we will expand on this by showing that length in $\mathbf{V}^{\wedge n}$ provides a good definition of volume in $\mathbf{V}.$

Definition 1: For any vectors $\mathbf{v}_1,\dots,\mathbf{v}_n,$ the corresponding parallelepiped is the subset of $\mathbf{V}$ defined by

$\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n) := \{ t_1\mathbf{v}_1 + \dots + t_n\mathbf{v}_n \colon 0 \leq t_i \leq 1\}.$

We define the volume of this parallelepiped by

$\mathrm{Vol} \mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n) := \| \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\|.$

According to Definition 1, we should interpret the length of the tensor $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$ to be the volume of the parallelepiped spanned by its factors. This makes sense, in that the length of $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n$ is zero if and only if $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ is a linearly dependent set in $\mathbf{V},$ which means that $\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n)$ lies in a subspace of $\mathbf{V}$ whose dimension strictly less than $n$. 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 $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ is a basis in $\mathbf{V}$ if and only if $\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n)>0.$

As an example, let us consider the case $n=1,$ i.e. $\mathbf{V}$ is a $1$-dimensional vector space. Let $\mathbf{v} \in \mathbf{V}.$ We then have that $\mathcal{P}(\mathbf{v})$ consists of all vectors of the form $w=t\mathbf{v}$ with $0 \leq t \leq 1,$ which can be visualized as the line segment joining $\mathbf{0}_\mathbf{V}$ to $\mathbf{v}$ in $\mathbf{V}.$ The volume of $\mathcal{P}(V)$ is then $\|\mathbf{v}\|,$ 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 $n=2,$ i.e. $\mathbf{V}$ is a $2$-dimensional Euclidean space. Let $\mathbf{v}_1,\mathbf{v}_2 \in \mathbf{V}.$ Then, $\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2)$ can be visualized as the set of all vectors in $\mathbf{V}$ which lie on or inside the parallelogram with vertices

$\mathbf{0}_\mathbf{V} = 0\mathbf{v}_1 + 0\mathbf{v}_2 \\ \mathbf{v}_1 = 1\mathbf{v}_1 + 0\mathbf{v}_2 \\ \mathbf{v}_2 = 0\mathbf{v}_1 + 1\mathbf{v}_2 \\ \mathbf{v}_1 + \mathbf{v}_2 = 1\mathbf{v}_1 + 1\mathbf{v}_2.$

According to our definition, the volume of $\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2)$ is the number

$\mathrm{Vol} \mathcal{P}(\mathbf{v}_1,\mathbf{v}_2) = \|\mathbf{v}_1 \wedge \mathbf{v}_2\|.$

Let us evaluate this number more explicitly. Let $E = \{\mathbf{e}_1,\mathbf{e}_2\}$ be an orthonormal basis of $\mathbf{V}.$ We then have

$\mathbf{v}_1 \wedge \mathbf{v}_2 = (\langle \mathbf{e}_1,\mathbf{v}_1\rangle \mathbf{e}_1 + \langle \mathbf{e}_2,\mathbf{v}_1\rangle \mathbf{e}_2)\wedge (\langle \mathbf{e}_1,\mathbf{v}_2\rangle \mathbf{e}_1 + \langle \mathbf{e}_2,\mathbf{v}_2\rangle \mathbf{e}_2) \\= \left( \langle \mathbf{e}_1,\mathbf{v}_1\rangle\langle \mathbf{e}_2,\mathbf{v}_2\rangle - \langle \mathbf{e}_1,\mathbf{v}_2\rangle\langle\mathbf{e}_2,\mathbf{v}_1\rangle\right)\mathbf{e}_1 \wedge \mathbf{e}_2,$

so that

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2) = |\langle \mathbf{e}_1,\mathbf{v}_1\rangle\langle \mathbf{e}_2,\mathbf{v}_2\rangle - \langle \mathbf{e}_1,\mathbf{v}_2\rangle\langle\mathbf{e}_2,\mathbf{v}_1\rangle| \|\mathbf{e}_1 \wedge \mathbf{e}_2\|.$

Since

$\|\mathbf{e}_1 \wedge \mathbf{e}_2\|^2 = \langle \mathbf{e}_1 \wedge \mathbf{e}_2,\mathbf{e}_1\wedge \mathbf{e}_2 \rangle = \langle \mathbf{e}_1,\mathbf{e}_1\rangle \langle \mathbf{e}_2,\mathbf{e}_2\rangle-\langle \mathbf{e}_1,\mathbf{e}_2\rangle \langle \mathbf{e}_2,\mathbf{e}_1\rangle = \|\mathbf{e}_1\|^2\|\mathbf{e}_2\|^2=1,$

we conclude that

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2) = |\langle \mathbf{e}_1,\mathbf{v}_1\rangle\langle \mathbf{e}_2,\mathbf{v}_2\rangle - \langle \mathbf{e}_1,\mathbf{v}_2\rangle\langle\mathbf{e}_2,\mathbf{v}_1\rangle|.$

There are a couple of things we can notice observe about this number. First, we can also write it as

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2) =| \langle \mathbf{e}_1 \wedge \mathbf{e}_2,\mathbf{v}_1 \wedge \mathbf{v}_2|;$

as we will see momentarily, this is because $\{\mathbf{e}_1\wedge \mathbf{e}_2\}$ is an orthonormal basis of $\mathbf{V}^{\wedge 2}.$ Second, if you are familiar with the definition of the determinant of a $2 \times 2$ matrix,

$\det \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = a_{11}a_{22}-a_{12}a_{21},$

it is apparent that $\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2)$ is equal to the absolute value of the determinant of the matrix

$\begin{bmatrix} \langle \mathbf{e}_1,\mathbf{v}_1\rangle & \langle \mathbf{e}_1,\mathbf{v}_2\rangle\\ \langle \mathbf{e}_2,\mathbf{v}_1\rangle & \langle \mathbf{e}_2,\mathbf{v}_2\rangle\end{bmatrix}$

whose columns are the coordinates of the vectors $\mathbf{v}_1,\mathbf{v}_2$ relative to the basis $E=\{\mathbf{e}_1.\mathbf{e}_2\}.$ 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 $F=\{\mathbf{f}_1,\mathbf{f}_2\}$ of $\mathbf{V}$? The same calculations would then have led us to the formula

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2) = |\langle \mathbf{f}_1,\mathbf{v}_1\rangle\langle \mathbf{f}_2,\mathbf{v}_2\rangle - \langle \mathbf{f}_1,\mathbf{v}_2\rangle\langle\mathbf{f}_2,\mathbf{v}_1\rangle|,$

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 $U \in \mathrm{Aut}\mathbf{V}$ uniquely determined by

$U\mathbf{e}_1 = \mathbf{f}_1,\ U\mathbf{e}_2 = \mathbf{f}_2.$

We then see that

$|\langle \mathbf{f}_1,\mathbf{v}_1\rangle\langle \mathbf{f}_2,\mathbf{v}_2\rangle - \langle \mathbf{f}_1,\mathbf{v}_2\rangle\langle\mathbf{f}_2,\mathbf{v}_1\rangle| = |\langle U\mathbf{e}_1,\mathbf{v}_1\rangle\langle U\mathbf{e}_2,\mathbf{v}_2\rangle - \langle U\mathbf{e}_1,\mathbf{v}_2\rangle\langle U\mathbf{e}_2,\mathbf{v}_1\rangle|\\=|\langle \mathbf{e}_1,U^{-1}\mathbf{v}_1\rangle\langle \mathbf{e}_2,U^{-1}\mathbf{v}_2\rangle - \langle \mathbf{e}_1,U^{-1}\mathbf{v}_2\rangle\langle \mathbf{e}_2,U^{-1}\mathbf{v}_1\rangle|,$

which is equivalent to the assertion that

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2) = \mathrm{Vol}\mathcal{P}(U^{-1}\mathbf{v}_1,U^{-1}\mathbf{v}_2).$

This makes perfect geometric sense: since orthogonal transformations preserve lengths and angles, the parallelogram $\mathcal{P}(U^{-1}\mathbf{v}_1,U^{-1}\mathbf{v}_2)$ is just a rotated and/or flipped copy of $\mathcal{P}(\mathbf{v}_1,\mathbf{v}_2),$ which naturally has the same area.

The key step in understanding the linear algebra underlying our definition of $n$-dimensional volume is finding an orthonormal basis of $\mathbf{V}^{\wedge d},$ the space of antisymmetric tensors of degree $d.$ We have solved this problem for $\mathbf{V}^{\otimes d},$ the space of all tensors of degree $d;$ back in Lecture 22, we showed that any orthonormal basis $E=\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ in $\mathbf{V}$ induces a corresponding orthonormal basis

$E^{\otimes d} = \{\mathbf{e}_{i(1)} \otimes \dots \otimes \mathbf{e}_{i(d)} \colon i \in \mathrm{Fun}(d,n)\}$

in $\mathbf{V}^{\otimes d},$ so that $\dim \mathbf{V}^{\otimes d} = n^d.$ To obtain the analogous result for $\mathbf{V}^{\wedge d},$ we need to introduce the subset of $\mathrm{Fun}(d,n)$ consisting of increasing functions:

$\mathrm{Inc}(d,n):=\{i \in \mathrm{Fun}(d,n) \colon i(1) < i(2) < \dots < i(d)\}.$

Theorem 2: For any $1 \leq d \leq n,$ the set

$E^{\wedge d} = \{\mathbf{e}_{i(1)} \wedge \dots \wedge \mathbf{e}_{i(d)} \colon i \in \mathrm{Inc}(d,n)\}$

is an orthonormal basis in $\mathbf{V}^{\wedge d}.$ For all $d>n,$ the only antisymmetric tensor of degree $d$ is the zero tensor, i.e. $\mathbf{V}^{\wedge d}=\{\mathbf{0}_{\mathbf{V}^{\otimes d}}\}.$

Proof: First we prove that $E^{\wedge d}$ spans $\mathbf{V}^{\wedge d}.$ To do this, it is sufficient to show that any simple tensor

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d$

is a linear combination of the tensors in $E^{\wedge d}.$ But this is clear from the multilinearity of the wedge product: we have

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_d \\ = \left( \sum_{i=1}^n \langle \mathbf{e}_i, \mathbf{v}_1\rangle \mathbf{e}_i\right) \wedge \dots \wedge \left( \sum_{i=1}^n \langle \mathbf{e}_i, \mathbf{v}_d\rangle \mathbf{e}_i\right) \\ = \sum\limits_{i \in \mathrm{Fun}(d,n)} \langle \mathbf{e}_{i(1)},\mathbf{v}_1\rangle \dots \mathbf{e}_{i(d)},\mathbf{v}_d\rangle \mathbf{e}_{i(1)} \wedge \dots \wedge \mathbf{e}_{i(d)}.$

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 $i \in \mathrm{Fun}(d,n)$ such that the numbers $i(1),\dots,i(d)$ are pairwise distinct. If $d > n,$ there are no such functions, and consequently $\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_{d} = \mathbf{0}. If$latex d \leq n,\$ then for any injection $i,$ the factors of the nonzero tensor $e_{i(1)} \wedge \dots \wedge \mathbf{e}_{i(d)}$ can be sorted so that the distinct indices are in increasing order; doing this will scale the original tensor by a factor of $\pm 1.$

Now we prove that $E^{\wedge d}$ is an orthonormal set in $\mathbf{V}^{\wedge d}$ (in particular, this will imply linear independence). Let $i,j \in \mathrm{Inc}(d,n)$ be distinct increasing functions. We then have that

$\langle \mathbf{e}_{i(1)} \wedge \dots \wedge \mathbf{e}_{i(d)},\mathbf{e}_{j(1)} \wedge \dots \wedge \mathbf{e}_{j(d)} = \sum\limits_{\sigma \in \mathrm{S}(d)} \mathrm{sgn}(\sigma) \langle \mathbf{e}_{i(1)},\mathbf{e}_{j\sigma(1)}\rangle \dots \langle \mathbf{e}_{i(d)},\mathbf{e}_{j\sigma(d)}\rangle.$

Since there is no permutation $\sigma \in \mathrm{S}(d)$ which can transform the list $i(1) < \dots < i(d)$ of distinct numbers into the different list $j(1) < \dots < j(d)$ of distinct numbers, every term in the sum is zero. This proves that $E^{\wedge d}$ is an orthogonal set in $\mathbf{V}^{\wedge d}.$ To prove that each element of $E^{\wedge d}$ is of unit length, repeat the above argument with $i=j.$ Then, the only nonzero term in the sum is that corresponding to the identity permutation:

$\langle \mathbf{e}_i(1),\mathbf{e}_{I(d)} \rangle \dots \langle \mathbf{e}_i(1),\mathbf{e}_{I(d)} \rangle = 1.$

— Q.E.D.

Corollary 1: The dimension of $\mathbf{V}^{\wedge d}$ is ${n \choose d}.$

Proof: Any increasing function $I(1) < \dots < I(d)$ corresponds to a unique cardinality $d$ subset $\{i(1),\dots,i(d)\}$ of $\{1,\dots,n\}$. By definition, the binomial coefficient ${n \choose d}$ is the number of such sets.

— Q.E.D.

Although it may not seem so, the most important consequence of Corollary 1 is that

$\mathbf{V}^{\wedge n} = {n \choose n} =1.$

Indeed, according to Theorem 1, the set $E^{\wedge n} =\{\omega\}$ consisting of the single tensor

$\omega = \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n$

is a basis in $\mathbf{V}^{\wedge n}.$ This means that every antisymmetric tensor of degree $n$ is a scalar multiple of $\omega,$ so that in particular for any vectors $\mathbf{v}_1,\dots,\mathbf{v}_n \in \mathbf{V}$ we have

$\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n = a \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n.$

for some scalar $a \in \mathbb{R},$ the value of which depends on the vectors $\mathbf{v}_1,\dots,\mathbf{v}_n.$ Indeed, since $\{\mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\}$ is an orthonormal basis in $\mathbf{V}^{\wedge n},$ we have

$a = \langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle.$

We thus have the following alternative description of volume.

Proposition 1: For any orthonormal basis $E=\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ in $\mathbf{V},$ we have

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n) = |\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle|.$

Proof: We will give two proofs, one computational and one conceptual.

Computational:

$\mathrm{Vol}\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n) \\=\|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\| \\ = \sqrt{\langle \mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle} \\ = \sqrt{\langle\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\rangle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\rangle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\rangle} \\ = \sqrt{\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\rangle^2\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\rangle} \\ = \sqrt{\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\rangle^2} \\ =|\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle|.$

Conceptual: by Cauchy-Schwarz,

$|\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle| \leq \|\mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\| \|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\|.$

We know that equality holds in the Cauchy-Schwarz inequality precisely when linear dependence holds, which is the case here because $\dim \mathbf{V}^{\wedge n}=1.$ Thus

$|\langle \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle| =\|\mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\| \|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\| = \|\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n\|.$

— Q.E.D.

Any tensor $\omega \in \mathbf{V}^{\wedge n}$ of the form

$\omega=\mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,$

where $E=\{\mathbf{e}_1,\dots,\mathbf{e}_n\}$ is an orthonormal basis in $\mathbf{V}$, is called a unit volume tensor for $\mathbf{V}.$ Indeed, for such a tensor we have $\|\omega\|=1,$ corresponding exactly to the fact that $\mathcal{P}(\mathbf{e}_1,\dots,\mathbf{e}_n)$ is an $n$-dimensional unit box. In fact, because $\mathbf{V}^{\wedge n}$ is $1$-dimensional, there are only two unit volume tensors: if $\omega \in \mathbf{V}^{\wedge n}$ satisfies $\|\omega\|=1,$ then the only other tensor in $\mathbf{V}^{\wedge n}$ with this property is $-\omega.$

Definition 3: A triple $(\mathbf{V},\langle \cdot,\cdot \rangle,\omega)$ consisting of a finite-dimensional vector space $\mathbf{V},$ a scalar product $\langle \cdot,\cdot \rangle$ on $\mathbf{V}$, and a unit volume tensor $\omega \in \mathbf{V}^{\wedge n},$ is called an oriented Euclidean space.

For example, consider the case of the Euclidean plane, i.e. $\mathbf{V}=\mathbb{R}^2$ with the standard scalar product. Giving the Euclidean plane an orientation means choosing one of the two tensors

$(1,0) \wedge (0,1) \quad\text{ or }\quad (0,1) \wedge (1,0).$

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 $\omega \in \mathbf{V}^{\wedge n},$ the corresponding determinant is the function

$\det \colon \underbrace{\mathbf{V} \times \dots \times \mathbf{V}}_{n \text{ copies}} \to \mathbb{R}$

defined by

$\det(\mathbf{v}_1,\dots,\mathbf{v}_n) = \langle \omega,\mathbf{v}_1 \wedge \dots \wedge \mathbf{v}_n \rangle.$

Thus $\det(\mathbf{v}_1,\dots,\mathbf{v}_n)$ is the oriented volume of the parallelepiped $\mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n),$ meaning that it coincides with $\mathrm{Vol} \mathcal{P}(\mathbf{v}_1,\dots,\mathbf{v}_n)$ up to a factor of $\pm 1.$ 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 $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ is a basis in $\mathbf{V}$ if and only if $\det(\mathbf{v}_1,\dots,\mathbf{v}_n) \neq 0.$

It is important to realize that the Euclidean space $\mathbf{V}$ actually supports two distinct determinant functions, one for each of the two unit volume tensors in the one-dimensional space $\mathbf{V}^{\wedge n}.$ For example, if $n=2$ and we have two orthonormal bases $E=\{\mathbf{e}_1,\mathbf{e}_2\}$ and $F=\{\mathbf{f}_1,\mathbf{f}_2\}$ then we have two corresponding determinant functions

$\det_E(\mathbf{v}_1,\mathbf{v}_2) = \langle \mathbf{e}_1 \wedge \mathbf{e}_2, \mathbf{v}_1 \wedge \mathbf{v}_2\rangle \quad\text{ and }\quad \det_F(\mathbf{v}_1,\mathbf{v}_2) = \langle \mathbf{f}_1 \wedge \mathbf{f}_2, \mathbf{v}_1 \wedge \mathbf{v}_2\rangle$

which may or may not coincide. For example, if

$\mathbf{f}_1=\mathbf{e}_2,\ \mathbf{f}_2=\mathbf{e}_1,$

then we have

$\det_F(\mathbf{v}_1,\mathbf{v}_2) = \langle \mathbf{f}_1 \wedge \mathbf{f}_2,\mathbf{v}_1 \wedge \mathbf{v}_2 \rangle = \langle \mathbf{e}_2 \wedge \mathbf{e}_1,\mathbf{v}_1\wedge \mathbf{v}_2\rangle=-\det_E(\mathbf{v}_1,\mathbf{v}_2).$

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.

## 1 Comment

1. Sirius says:

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.