Alternating Tensors
Among all multilinear functionals on a vector space, two families behave predictably when their arguments are permuted.
One family is unchanged by every permutation; the other records the permutation through a single sign.
We have already studied the first — the symmetric tensors, whose values are insensitive to the order of their arguments.
Here we develop the second, which underlies the theory of integration and the exterior calculus on smooth manifolds:
the tensors that change sign under every transposition of arguments.
Throughout this page, \(V\) denotes a finite-dimensional real vector space, and we work with
covariant \(k\)-tensors
on \(V\), that is, real-valued multilinear functions of \(k\) arguments from \(V\). When a permutation \(\sigma\) of \(\{1, \dots, k\}\)
acts on such a tensor, we write \({}^{\sigma}\alpha\) for the tensor obtained by permuting the arguments,
\[
({}^{\sigma}\alpha)(v_1, \dots, v_k) = \alpha\bigl(v_{\sigma(1)}, \dots, v_{\sigma(k)}\bigr),
\]
and \(\operatorname{sgn}\sigma \in \{+1, -1\}\) for the sign of the permutation, equal to \(+1\) when \(\sigma\) decomposes
into an even number of transpositions and \(-1\) otherwise.
Definition: Alternating Tensor
A covariant \(k\)-tensor \(\alpha\) on \(V\) is alternating if its value changes sign whenever two of its arguments
are interchanged: for all \(v_1, \dots, v_k \in V\) and every pair of indices \(i \neq j\),
\[
\alpha(v_1, \dots, v_i, \dots, v_j, \dots, v_k) = -\,\alpha(v_1, \dots, v_j, \dots, v_i, \dots, v_k).
\]
The space of alternating covariant \(k\)-tensors on \(V\) is denoted \(\Lambda^k(V^*)\).
Its elements are also called exterior \(k\)-forms or \(k\)-covectors.
Since every permutation is a composition of transpositions, the defining condition propagates:
applying a permutation \(\sigma\) to the arguments of an alternating tensor multiplies its value
by \(\operatorname{sgn}\sigma\), so that \({}^{\sigma}\alpha = (\operatorname{sgn}\sigma)\,\alpha\).
Two boundary cases are immediate. Every \(0\)-tensor (a scalar) and every \(1\)-tensor (a single covector) is vacuously alternating,
since neither admits a pair of arguments to interchange; thus \(\Lambda^0(V^*) = \mathbb{R}\) and \(\Lambda^1(V^*) = V^*\).
The sign-reversal condition has two equivalent reformulations that are often more convenient in practice.
One detects alternation through linear dependence of the arguments; the other through their coincidence.
The following result establishes that all three descriptions coincide.
Theorem: Characterizations of Alternating Tensors
Let \(\alpha\) be a covariant \(k\)-tensor on a finite-dimensional vector space \(V\). The following are equivalent.
- \(\alpha\) is alternating.
- \(\alpha(v_1, \dots, v_k) = 0\) whenever the \(k\)-tuple \((v_1, \dots, v_k)\) is linearly dependent.
- \(\alpha\) gives the value zero whenever two of its arguments are equal.
Proof:
We show (1) \(\Rightarrow\) (3), (2) \(\Rightarrow\) (3), and finally (3) \(\Rightarrow\) (1) and (3) \(\Rightarrow\) (2), which
together close the cycle.
(1) \(\Rightarrow\) (3).
Suppose \(\alpha\) is alternating and two arguments coincide, say the \(i\)th and \(j\)th equal a common vector \(w\)
with \(i \neq j\). Interchanging these two arguments leaves the input list unchanged, so the value is unaffected;
yet alternation forces it to change sign. Hence the value equals its own negative, and therefore vanishes.
(2) \(\Rightarrow\) (3).
If two arguments are equal, the \(k\)-tuple is linearly dependent, so condition (2) gives the value zero directly.
(3) \(\Rightarrow\) (1).
Assume \(\alpha\) vanishes whenever two arguments coincide. Fix a pair of indices \(i \neq j\) and consider,
for arbitrary \(v_1, \dots, v_k\), the result of placing the sum \(v_i + v_j\) in both the \(i\)th and \(j\)th slots
while keeping the others fixed. The two equal arguments force
\[
0 = \alpha(\dots, v_i + v_j, \dots, v_i + v_j, \dots).
\]
Expanding by multilinearity in these two slots produces four terms:
\[
\begin{align*}
0 = {}&\alpha(\dots, v_i, \dots, v_i, \dots) + \alpha(\dots, v_i, \dots, v_j, \dots) \\\\
&+ \alpha(\dots, v_j, \dots, v_i, \dots) + \alpha(\dots, v_j, \dots, v_j, \dots).
\end{align*}
\]
The first and last terms each have a repeated argument and so vanish by hypothesis. What remains is
\[
\alpha(\dots, v_i, \dots, v_j, \dots) = -\,\alpha(\dots, v_j, \dots, v_i, \dots),
\]
which is exactly the sign-reversal condition for the transposed pair.
Since \(i\) and \(j\) were arbitrary, \(\alpha\) is alternating.
(3) \(\Rightarrow\) (2).
Suppose again that \(\alpha\) vanishes on repeated arguments, and let \((v_1, \dots, v_k)\) be linearly dependent.
Then some \(v_m\) is a linear combination of the others, say \(v_m = \sum_{l \neq m} a^l v_l\).
Substituting this expression into the \(m\)th slot and expanding by linearity there yields a sum of terms,
each of which is \(a^l\) times the value of \(\alpha\) on a list in which \(v_l\) occupies both the \(l\)th
and \(m\)th slots. Every such term has a repeated argument, hence vanishes, and so \(\alpha(v_1, \dots, v_k) = 0\).
Characterization (2) exposes a structural consequence that will govern the entire theory: an alternating \(k\)-tensor annihilates
every linearly dependent list of arguments. In an \(n\)-dimensional space, any list of more than \(n\) vectors
is automatically dependent, so every alternating tensor of degree greater than \(n\) must vanish identically.
This degree ceiling — invisible for symmetric tensors, which exist in every degree — is the first sign
that the spaces \(\Lambda^k(V^*)\) form a finite, tightly constrained hierarchy. We return to it once we have built the tools
to count dimensions.
The Alternation Operator
Not every covariant tensor is alternating, but every one has an alternating part. Just as the
symmetrization operator
extracts the symmetric part of a tensor by averaging over all permutations of its arguments,
there is a companion operator that extracts the alternating part by averaging the signed permutations.
It is the projection that makes the alternating tensors into a direct summand of the full tensor space, and
it will be the engine behind the wedge product in the next stage of the theory.
Definition: The Alternation Operator
Let \(\alpha\) be a covariant \(k\)-tensor on \(V\). The alternation of \(\alpha\) is the covariant \(k\)-tensor
\[
\operatorname{Alt}\alpha = \frac{1}{k!} \sum_{\sigma \in S_k} (\operatorname{sgn}\sigma)\,{}^{\sigma}\alpha,
\]
where the sum runs over the
symmetric group
\(S_k\) of permutations of \(\{1, \dots, k\}\). Evaluated on arguments, this reads
\[
(\operatorname{Alt}\alpha)(v_1, \dots, v_k) = \frac{1}{k!} \sum_{\sigma \in S_k} (\operatorname{sgn}\sigma)\,\alpha\bigl(v_{\sigma(1)}, \dots, v_{\sigma(k)}\bigr).
\]
The normalizing factor \(1/k!\) is chosen so that the operator fixes tensors that are already alternating;
without it, \(\operatorname{Alt}\) would scale them by \(k!\). The next examples make the definition concrete in low degree.
Example: Alternation in Low Degrees
For \(k = 1\), the group \(S_1\) is trivial, so \(\operatorname{Alt}\alpha = \alpha\): every \(1\)-tensor
is already alternating, as noted above. For a \(2\)-tensor \(\beta\),
the group \(S_2\) has the identity and one transposition, giving
\[
(\operatorname{Alt}\beta)(v, w) = \tfrac{1}{2}\bigl(\beta(v, w) - \beta(w, v)\bigr).
\]
For a \(3\)-tensor \(\gamma\), summing the signed permutations of \(S_3\) yields
\[
\begin{align*}
(\operatorname{Alt}\gamma)(u, v, w) = \tfrac{1}{6}\bigl(
&\gamma(u, v, w) + \gamma(v, w, u) + \gamma(w, u, v) \\\\
&- \gamma(v, u, w) - \gamma(u, w, v) - \gamma(w, v, u)\bigr),
\end{align*}
\]
with a plus sign on the three even permutations and a minus sign on the three odd ones.
The operator earns its name through the following properties, which mirror those of symmetrization.
The first says that \(\operatorname{Alt}\) always lands in \(\Lambda^k(V^*)\); the second says that it acts as the identity there,
and on no larger subspace. Together they identify \(\operatorname{Alt}\) as a projection of the tensor space onto its alternating part.
Theorem: Properties of the Alternation Operator
Let \(\alpha\) be a covariant \(k\)-tensor on a finite-dimensional vector space \(V\).
- \(\operatorname{Alt}\alpha\) is alternating.
- \(\operatorname{Alt}\alpha = \alpha\) if and only if \(\alpha\) is alternating.
Proof:
Part (1).
We show that applying an arbitrary transposition \(\tau \in S_k\) to the arguments of \(\operatorname{Alt}\alpha\)
reverses its sign; by the characterization established in the previous section, this makes \(\operatorname{Alt}\alpha\) alternating.
By definition,
\[
\begin{align*}
{}^{\tau}(\operatorname{Alt}\alpha)
&= \frac{1}{k!} \sum_{\sigma \in S_k} (\operatorname{sgn}\sigma)\,{}^{\tau}({}^{\sigma}\alpha) \\\\
&= \frac{1}{k!} \sum_{\sigma \in S_k} (\operatorname{sgn}\sigma)\,{}^{\tau\sigma}\alpha,
\end{align*}
\]
using that permuting arguments composes as
\({}^{\tau}({}^{\sigma}\alpha) = {}^{\tau\sigma}\alpha\). This composition rule follows directly from the definition:
writing \(w_i = v_{\tau(i)}\), one has
\[
\begin{align*}
\bigl({}^{\tau}({}^{\sigma}\alpha)\bigr)(v_1, \dots, v_k)
&= ({}^{\sigma}\alpha)(w_1, \dots, w_k) \\\\
&= \alpha(w_{\sigma(1)}, \dots, w_{\sigma(k)}) \\\\
&= \alpha\bigl(v_{\tau(\sigma(1))}, \dots, v_{\tau(\sigma(k))}\bigr),
\end{align*}
\]
which is \({}^{\tau\sigma}\alpha\) evaluated at \((v_1, \dots, v_k)\).
Substitute \(\rho = \tau\sigma\); as \(\sigma\) ranges over \(S_k\) so does \(\rho\),
and \(\operatorname{sgn}\sigma = \operatorname{sgn}(\tau^{-1}\rho) = (\operatorname{sgn}\tau)(\operatorname{sgn}\rho)\).
Since \(\tau\) is a transposition, \(\operatorname{sgn}\tau = -1\), so
\[
\begin{align*}
{}^{\tau}(\operatorname{Alt}\alpha)
&= \frac{1}{k!} \sum_{\rho \in S_k} (\operatorname{sgn}\tau)(\operatorname{sgn}\rho)\,{}^{\rho}\alpha \\\\
&= (\operatorname{sgn}\tau)\,\operatorname{Alt}\alpha \\\\
&= -\operatorname{Alt}\alpha.
\end{align*}
\]
Thus \(\operatorname{Alt}\alpha\) reverses sign under every transposition and is therefore alternating.
Part (2).
If \(\operatorname{Alt}\alpha = \alpha\), then \(\alpha\) is alternating by Part (1).
Conversely, suppose \(\alpha\) is alternating.
Then \({}^{\sigma}\alpha = (\operatorname{sgn}\sigma)\,\alpha\) for every \(\sigma \in S_k\), so each term of the sum equals
\[
(\operatorname{sgn}\sigma)\,{}^{\sigma}\alpha = (\operatorname{sgn}\sigma)(\operatorname{sgn}\sigma)\,\alpha = \alpha,
\]
because \((\operatorname{sgn}\sigma)^2 = 1\). The sum therefore has \(k!\) identical terms, and the factor \(1/k!\) returns
\[
\operatorname{Alt}\alpha = \frac{1}{k!} \sum_{\sigma \in S_k} \alpha = \alpha.
\]
Applying \(\operatorname{Alt}\) twice therefore changes nothing after the first application:
since \(\operatorname{Alt}\alpha\) is alternating, Part (2) gives
\(\operatorname{Alt}(\operatorname{Alt}\alpha) = \operatorname{Alt}\alpha\).
The operator is idempotent — a genuine projection of the space of covariant \(k\)-tensors
onto the subspace \(\Lambda^k(V^*)\). With a reliable way to produce alternating tensors in hand, we now construct
an explicit basis for the space they fill.
Elementary Alternating Tensors
To build a basis for \(\Lambda^k(V^*)\), we manufacture a distinguished alternating tensor
from each list of \(k\) basis covectors. The construction packages a determinant: evaluated on \(k\) vectors,
the elementary tensor returns the determinant of the matrix of their selected components.
Determinants are the prototypical alternating functions of their columns, so alternation is built in from the start.
Throughout, fix a basis \((E_1, \dots, E_n)\) for \(V\) with dual basis \((\varepsilon^1, \dots, \varepsilon^n)\) for \(V^*\), so that
\(\varepsilon^i(E_j) = \delta^i_j\).
We index our constructions by
multi-indices
\(I = (i_1, \dots, i_k)\) with each entry in \(\{1, \dots, n\}\).
Definition: Elementary Alternating Tensor
Let \(I = (i_1, \dots, i_k)\) be a multi-index of length \(k\) with entries in \(\{1, \dots, n\}\).
The associated elementary alternating tensor \(\varepsilon^I = \varepsilon^{i_1 \cdots i_k} \in \Lambda^k(V^*)\) is defined by
\[
\varepsilon^I(v_1, \dots, v_k) = \det
\begin{pmatrix}
\varepsilon^{i_1}(v_1) & \cdots & \varepsilon^{i_1}(v_k) \\\\
\vdots & \ddots & \vdots \\\\
\varepsilon^{i_k}(v_1) & \cdots & \varepsilon^{i_k}(v_k)
\end{pmatrix}.
\]
Equivalently, if \(\mathbf{v}\) is the \(n \times k\) matrix whose columns are the component vectors of \(v_1, \dots, v_k\)
in the basis \((E_i)\), then \(\varepsilon^I(v_1, \dots, v_k)\) is the determinant of the \(k \times k\) submatrix formed
from rows \(i_1, \dots, i_k\) of \(\mathbf{v}\).
Because the determinant changes sign whenever two of its columns are interchanged, \(\varepsilon^I\) is
alternating in its arguments — its degree-\(k\) multilinearity and sign behavior are inherited wholesale from the determinant.
This justifies the name and the membership \(\varepsilon^I \in \Lambda^k(V^*)\).
Example: Elementary Tensors on \(\mathbb{R}^3\)
Take \(V = \mathbb{R}^3\) with the standard dual basis \((\varepsilon^1, \varepsilon^2, \varepsilon^3)\), where \(\varepsilon^i\)
reads off the \(i\)th coordinate. Then for vectors \(v, w\),
\[
\varepsilon^{13}(v, w) = \det
\begin{pmatrix}
v^1 & w^1 \\\\
v^3 & w^3
\end{pmatrix}
= v^1 w^3 - w^1 v^3,
\]
the signed area of the projection of the parallelogram onto the \((1,3)\)-coordinate plane. At full degree,
\[
\varepsilon^{123}(u, v, w) = \det(u, v, w),
\]
the ordinary determinant of the three component vectors.
Evaluating an elementary tensor on a list of basis vectors produces a clean combinatorial answer, which is most efficiently
expressed by extending the Kronecker delta to multi-indices.
Given multi-indices \(I = (i_1, \dots, i_k)\) and \(J = (j_1, \dots, j_k)\), set
\[
\delta^I_J = \det
\begin{pmatrix}
\delta^{i_1}_{j_1} & \cdots & \delta^{i_1}_{j_k} \\\\
\vdots & \ddots & \vdots \\\\
\delta^{i_k}_{j_1} & \cdots & \delta^{i_k}_{j_k}
\end{pmatrix}.
\]
This generalized symbol evaluates to \(+1\) or \(-1\) when \(I\) and \(J\) have no repeated entries and \(J\) is a permutation of \(I\)
— recording the sign of that permutation — and to \(0\) otherwise. The following lemma collects the properties that
make \(\varepsilon^I\) the right building block.
Theorem: Properties of Elementary Alternating Tensors
Let \((E_i)\) be a basis for \(V\), let \((\varepsilon^i)\) be its dual basis, and let \(\varepsilon^I\)
be defined as above for a multi-index \(I\) of length \(k\).
- If \(I\) has a repeated index, then \(\varepsilon^I = 0\).
- If \(J = I_\sigma\) is obtained from \(I\) by a permutation \(\sigma \in S_k\),
then \(\varepsilon^J = (\operatorname{sgn}\sigma)\,\varepsilon^I\).
- The result of evaluating \(\varepsilon^I\) on a sequence of basis vectors is
\[
\varepsilon^I(E_{j_1}, \dots, E_{j_k}) = \delta^I_J,
\]
where \(J = (j_1, \dots, j_k)\).
Proof:
Part (1).
If \(I\) has a repeated index, then the defining determinant has two identical rows, and a determinant with two equal rows is zero.
Hence \(\varepsilon^I\) is the zero tensor.
Part (2).
Permuting the entries of \(I\) by \(\sigma\) permutes the rows of the defining determinant by \(\sigma\).
Interchanging two rows of a determinant reverses its sign, so applying \(\sigma\) multiplies the determinant
by \(\operatorname{sgn}\sigma\). Therefore \(\varepsilon^J = (\operatorname{sgn}\sigma)\,\varepsilon^I\).
Part (3).
Evaluating on basis vectors, the \((p, q)\)-entry of the defining matrix is
\(\varepsilon^{i_p}(E_{j_q}) = \delta^{i_p}_{j_q}\), which is precisely the \((p, q)\)-entry of the matrix defining \(\delta^I_J\).
The two determinants therefore agree, giving \(\varepsilon^I(E_{j_1}, \dots, E_{j_k}) = \delta^I_J\).
Part (1) shows that an elementary tensor vanishes unless its multi-index has distinct entries, and Part (2)
shows that reordering those entries only flips a sign. So nothing is lost, and no redundancy introduced, by restricting attention
to increasing multi-indices — those with \(i_1 \lt \cdots \lt i_k\)
— one representative from each set of distinct indices. We record this convention for the basis construction that follows.
Definition: Increasing Multi-index
A multi-index \(I = (i_1, \dots, i_k)\) is increasing if \(i_1 \lt i_2 \lt \cdots \lt i_k\).
A sum taken only over increasing multi-indices of length \(k\) is written with a primed summation sign,
\[
{\sum_I}'\, a_I\, \varepsilon^I = \sum_{\{I \,:\, i_1 \lt \cdots \lt i_k\}} a_I\, \varepsilon^I,
\]
to distinguish it from a sum over all multi-indices.
Part (3) of the lemma is the key computational fact: an increasing elementary tensor \(\varepsilon^I\)
evaluates to \(1\) on the matching increasing list of basis vectors \((E_{i_1}, \dots, E_{i_k})\)
and to \(0\) on every other increasing list. This is exactly the orthogonality relation a basis and its dual basis satisfy —
and it is what makes the elementary alternating tensors a basis for \(\Lambda^k(V^*)\), the result we establish next.
A Basis for the Space of Alternating Tensors
We can now assemble the pieces. The increasing elementary tensors are linearly independent and span \(\Lambda^k(V^*)\), so
they form a basis. Counting them gives the dimension of the space — a binomial coefficient — and exposes the degree ceiling promised
at the outset: above the dimension of \(V\), no nonzero alternating tensors survive.
Theorem: A Basis for the Alternating Tensors
Let \(V\) be an \(n\)-dimensional vector space with basis \((E_i)\) and dual basis \((\varepsilon^i)\).
For each positive integer \(k \leq n\), the collection
\[
\bigl\{\, \varepsilon^I : I \text{ is an increasing multi-index of length } k \,\bigr\}
\]
is a basis for \(\Lambda^k(V^*)\). Consequently
\[
\dim \Lambda^k(V^*) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!}.
\]
If \(k \gt n\), then \(\Lambda^k(V^*) = \{0\}\).
Proof:
The claim for \(k \gt n\) is immediate from the characterization of alternating tensors:
any \(k\)-tuple of vectors in an \(n\)-dimensional space with \(k \gt n\) is linearly dependent,
so every alternating \(k\)-tensor vanishes on every input and is therefore the zero tensor.
We now treat \(k \leq n\), showing that the increasing elementary tensors span \(\Lambda^k(V^*)\)
and are linearly independent.
Spanning.
Let \(\alpha \in \Lambda^k(V^*)\) be arbitrary. For each increasing multi-index \(I = (i_1, \dots, i_k)\), define the scalar
\[
\alpha_I = \alpha(E_{i_1}, \dots, E_{i_k}),
\]
and form the candidate \(\beta = {\sum_I}'\, \alpha_I\, \varepsilon^I\), the primed sum running over increasing multi-indices.
We claim \(\beta = \alpha\). Both are alternating \(k\)-tensors, so by multilinearity it suffices
to check that they agree on every list of basis vectors \((E_{j_1}, \dots, E_{j_k})\); and since both vanish
when two arguments coincide, we may further assume the indices \(j_1, \dots, j_k\) are distinct. An alternating tensor
is determined on a list of distinct basis vectors by its value on the increasing rearrangement of that list, up to the sign
of the rearranging permutation, so it is enough to verify agreement on increasing lists \(J = (j_1, \dots, j_k)\). For such a \(J\),
Part (3) of the elementary-tensor lemma gives \(\varepsilon^I(E_{j_1}, \dots, E_{j_k}) = \delta^I_J\),
which equals \(1\) when the increasing multi-indices \(I\) and \(J\) coincide and \(0\) otherwise. Hence
\[
\begin{align*}
\beta(E_{j_1}, \dots, E_{j_k})
&= {\sum_I}'\, \alpha_I\, \delta^I_J \\\\
&= \alpha_J \\\\
&= \alpha(E_{j_1}, \dots, E_{j_k}).
\end{align*}
\]
So \(\alpha\) and \(\beta\) agree on all increasing lists of basis vectors, hence on all lists, hence as tensors.
Therefore \(\alpha = {\sum_I}'\, \alpha_I\, \varepsilon^I\), and the increasing elementary tensors span \(\Lambda^k(V^*)\).
Linear independence.
Suppose a vanishing combination \({\sum_I}'\, c_I\, \varepsilon^I = 0\) holds for scalars \(c_I\), with the sum over increasing
multi-indices. Evaluate both sides on an arbitrary increasing list of basis vectors \((E_{j_1}, \dots, E_{j_k})\),
with \(J = (j_1, \dots, j_k)\). By Part (3) of the lemma, every term vanishes except the one with \(I = J\),
which contributes \(c_J\). Thus \(c_J = 0\). Since \(J\) was an arbitrary increasing multi-index, all coefficients vanish,
and the increasing elementary tensors are linearly independent.
The collection is therefore a basis. Its cardinality is the number of increasing multi-indices of length \(k\) drawn
from \(\{1, \dots, n\}\), which is the number of \(k\)-element subsets of an \(n\)-element set, namely \(\binom{n}{k}\).
This is the dimension of \(\Lambda^k(V^*)\).
The top degree \(k = n\) is special. There is exactly one increasing multi-index of length \(n\), namely \((1, 2, \dots, n)\),
so \(\Lambda^n(V^*)\) is one-dimensional, spanned by \(\varepsilon^{1 \cdots n}\).
By definition this tensor sends a list of vectors to the determinant of their component matrix;
on \(\mathbb{R}^n\) with the standard basis, \(\varepsilon^{1 \cdots n}\) is precisely the determinant function.
Every alternating \(n\)-tensor is a scalar multiple of it.
This one-dimensionality has a consequence that reaches well beyond the present chapter: it pins down exactly
how a top-degree alternating tensor responds to a linear map applied to all of its arguments.
The answer is the determinant of the map — the same determinant that governs how volumes scale,
and the mechanism by which the change-of-variables factor will later enter the integration of differential forms.
Theorem: The Determinant from Top-Degree Alternation
Let \(V\) be an \(n\)-dimensional vector space, let \(\omega \in \Lambda^n(V^*)\)
be an alternating \(n\)-tensor, and let \(T : V \to V\) be a linear map.
Then for all \(v_1, \dots, v_n \in V\),
\[
\omega(Tv_1, \dots, Tv_n) = (\det T)\,\omega(v_1, \dots, v_n).
\]
Proof:
Define a map \(\omega_T(v_1, \dots, v_n) = \omega(Tv_1, \dots, Tv_n)\).
Since \(T\) is linear and \(\omega\) is multilinear and alternating, \(\omega_T\) is again an alternating \(n\)-tensor:
linearity in each slot is inherited through \(T\), and interchanging two arguments of \(\omega_T\)
interchanges two arguments of \(\omega\), reversing the sign. Thus \(\omega_T \in \Lambda^n(V^*)\).
Because \(\Lambda^n(V^*)\) is one-dimensional, \(\omega_T\) is a scalar multiple of \(\omega\),
say \(\omega_T = c\,\omega\) for some \(c \in \mathbb{R}\) depending on \(T\) but not on the arguments.
To identify \(c\), evaluate on the basis. Writing \(TE_j = \sum_i T^i_j\, E_i\) for the matrix \((T^i_j)\) of \(T\)
in the basis \((E_i)\), and using that \(\varepsilon^{1 \cdots n}\) is the determinant of the component matrix of its arguments,
\[
\begin{align*}
\omega_T(E_1, \dots, E_n)
&= \omega(TE_1, \dots, TE_n) \\\\
&= (\det T)\,\omega(E_1, \dots, E_n),
\end{align*}
\]
since the component matrix of \((TE_1, \dots, TE_n)\) is exactly \((T^i_j)\), whose determinant is \(\det T\).
Comparing with \(\omega_T(E_1, \dots, E_n) = c\,\omega(E_1, \dots, E_n)\) and noting that a nonzero \(\omega\) satisfies \(\omega(E_1, \dots, E_n) \neq 0\)
— since \(\omega = c_0\,\varepsilon^{1 \cdots n}\) with \(c_0 \neq 0\) and \(\varepsilon^{1 \cdots n}(E_1, \dots, E_n) = 1\), while the identity is trivially true for \(\omega = 0\) — we conclude \(c = \det T\).
Hence \(\omega_T = (\det T)\,\omega\), which is the claimed identity.
This identity is the algebraic seed of the determinant's geometric meaning.
A nonzero \(n\)-covector assigns a signed scale to each ordered \(n\)-tuple of vectors,
and the theorem says a linear map rescales every such assignment by the single number \(\det T\).
When these tensors are allowed to vary smoothly from point to point across a manifold — becoming the top-degree differential
forms of the next stage — this same factor reappears as the Jacobian determinant in the transformation of integrals.
Alternating Tensors as Signed Volume
The determinant identity makes precise an interpretation worth carrying forward: an alternating tensor measures signed volume.
Consider the parallelepiped spanned by vectors \(v_1, \dots, v_k\),
\[
P(v_1, \dots, v_k) = \{\, c_1 v_1 + \cdots + c_k v_k : 0 \leq c_l \leq 1 \,\},
\]
a parallelogram when \(k = 2\), an ordinary box when \(k = 3\). On \(\mathbb{R}^n\) at full degree \(k = n\),
the elementary tensor \(\varepsilon^{1 \cdots n}\) returns the determinant of the component matrix of its arguments,
and the absolute value of that determinant is exactly the \(n\)-dimensional volume of \(P(v_1, \dots, v_n)\).
The sign records orientation: positive when \((v_1, \dots, v_n)\) is ordered like the standard basis, negative
when the ordering is reversed. So \(\varepsilon^{1 \cdots n}\) is signed volume, and the theorem above
says a linear map \(T\) scales every signed volume by \(\det T\) — the classical statement that \(\det T\)
is the volume-scaling factor of \(T\), now read off from a one-dimensional space of top-degree tensors.
At intermediate degree the reading is the same, applied to a projection. The tensor \(\varepsilon^I\)
with \(I = (i_1, \dots, i_k)\) returns the signed \(k\)-dimensional volume of the parallelepiped obtained
by keeping only the coordinates \(i_1, \dots, i_k\) — the signed volume of the shadow that \(P(v_1, \dots, v_k)\)
casts on the corresponding coordinate subspace. The earlier value \(\varepsilon^{13}(v, w) = v^1 w^3 - w^1 v^3\)
is precisely the signed area of the shadow in the \((1,3)\)-plane.
This is the picture that motivates integrating alternating tensor fields over manifolds: each one
assigns an infinitesimal signed volume to a small spanning frame, and integration sums those volumes.
The Wedge Product
The symmetric tensors carry a product of their own: the symmetric product, formed by symmetrizing a tensor product.
Alternating tensors admit a parallel construction. Given an alternating \(k\)-tensor and an alternating \(l\)-tensor,
we form their tensor product and then alternate the result, obtaining an alternating \((k+l)\)-tensor.
This operation — the wedge product — turns the alternating tensors of all degrees into a single algebra,
and it is the operation through which differential forms acquire their computational power.
Recall that the tensor product of a \(k\)-tensor \(\omega\) and an \(l\)-tensor \(\eta\) is the \((k+l)\)-tensor
\[
(\omega \otimes \eta)(v_1, \dots, v_{k+l}) = \omega(v_1, \dots, v_k)\,\eta(v_{k+1}, \dots, v_{k+l}),
\]
extending the
tensor product of covectors.
This product is generally neither symmetric nor alternating; to land back in \(\Lambda^{k+l}(V^*)\) we apply the
alternation operator.
Definition: The Wedge Product
Let \(\omega \in \Lambda^k(V^*)\) and \(\eta \in \Lambda^l(V^*)\).
Their wedge product (or exterior product) is the alternating \((k+l)\)-tensor
\[
\omega \wedge \eta = \frac{(k+l)!}{k!\,l!}\,\operatorname{Alt}(\omega \otimes \eta).
\]
Written out on vectors, this is
\[
(\omega \wedge \eta)(v_1, \dots, v_{k+l})
= \frac{1}{k!\,l!} \sum_{\sigma \in S_{k+l}} (\operatorname{sgn}\sigma)\,\omega\bigl(v_{\sigma(1)}, \dots, v_{\sigma(k)}\bigr)\,\eta\bigl(v_{\sigma(k+1)}, \dots, v_{\sigma(k+l)}\bigr).
\]
The combinatorial coefficient \((k+l)!/(k!\,l!)\) is not cosmetic; it is precisely what is needed to make the elementary alternating tensors
multiply without clutter. The two forms of the definition agree because \(\operatorname{Alt}(\omega \otimes \eta)\)
carries the factor \(1/(k+l)!\), which cancels against the numerator, leaving \(1/(k!\,l!)\) in front of the sum.
The payoff of this normalization is the following clean rule for elementary tensors: the wedge product of \(\varepsilon^I\)
and \(\varepsilon^J\) is the elementary tensor indexed by the concatenation of \(I\) and \(J\). This is the property that
will let us read off wedge products from multi-indices alone.
Theorem: Wedge Products of Elementary Tensors
Let \((\varepsilon^i)\) be the dual basis of a basis \((E_i)\) for \(V\).
For any multi-indices \(I = (i_1, \dots, i_k)\) and \(J = (j_1, \dots, j_l)\), the wedge product of the
elementary alternating tensors
\(\varepsilon^I\) and \(\varepsilon^J\) is
\[
\varepsilon^I \wedge \varepsilon^J = \varepsilon^{IJ},
\]
where \(IJ = (i_1, \dots, i_k, j_1, \dots, j_l)\) is the concatenated multi-index of length \(k+l\).
Proof:
Both sides are alternating \((k+l)\)-tensors. Since an alternating tensor is determined by its values on
increasing lists of basis vectors,
it suffices to check that they agree on every sequence of basis vectors \((E_{p_1}, \dots, E_{p_{k+l}})\).
Write \(P = (p_1, \dots, p_{k+l})\). We consider the possible forms of \(P\).
Case 1: \(P\) has a repeated index.
Then both sides vanish — the right side by the property of \(\varepsilon^{IJ}\) on a repeated argument,
the left side because \(\varepsilon^I \wedge \varepsilon^J\), being alternating, also vanishes on a list
with a repeated basis vector.
Case 2: \(P\) contains an index appearing in neither \(I\) nor \(J\).
Then \(\varepsilon^{IJ}(E_{p_1}, \dots) = 0\), since the multi-index \(P\) is not a permutation of \(IJ\).
On the left, every term of the alternation sum evaluates either \(\varepsilon^I\) or \(\varepsilon^J\)
on a sub-list of basis vectors that is not a permutation of \(I\) or of \(J\) respectively, so each term vanishes;
hence the left side is zero as well.
Case 3: \(P\) is a permutation of \(IJ\) with no repeats.
Reordering the arguments by a permutation multiplies both sides by the same sign,
so it suffices to treat the case \(P = IJ\). Evaluating, the right side is
\(\varepsilon^{IJ}(E_{i_1}, \dots, E_{i_k}, E_{j_1}, \dots, E_{j_l}) = 1\).
On the left, among the \((k+l)!\) signed terms of the alternation, the only nonzero contributions come from permutations
\(\sigma\) that send the first \(k\) slots to a permutation of the positions carrying \(I\) and the last \(l\) slots
to a permutation of those carrying \(J\); the combinatorics, together with the coefficient \(1/(k!\,l!)\), collapse the sum to \(1\).
Hence both sides equal \(1\), completing the verification.
Iterating this rule, a wedge of several elementary tensors concatenates all their indices:
\[
\varepsilon^{i_1} \wedge \cdots \wedge \varepsilon^{i_k} = \varepsilon^{i_1 \cdots i_k} = \varepsilon^I.
\]
In particular every basis element of \(\Lambda^k(V^*)\) is a wedge of \(k\) basis covectors, which is
why the wedge product determines the entire algebraic structure once its behavior on covectors is known.
We turn next to the formal properties that make this structure an algebra.
Properties of the Wedge Product
The wedge product satisfies the axioms one expects of a product — it is bilinear and associative — together
with one notable departure from ordinary multiplication: it is anticommutative, reversing sign according
to the degrees of its factors. These properties, collected below, are enough to compute any wedge product mechanically,
and they recur unchanged when the construction is transported to differential forms.
Theorem: Properties of the Wedge Product
Let \(\omega, \omega', \eta, \eta', \xi\) be alternating tensors on a finite-dimensional vector space \(V\),
with \(\omega \in \Lambda^k(V^*)\) and \(\eta \in \Lambda^l(V^*)\) where degrees are relevant.
- Bilinearity. For scalars \(a, a' \in \mathbb{R}\),
\[
\begin{align*}
(a\omega + a'\omega') \wedge \eta &= a(\omega \wedge \eta) + a'(\omega' \wedge \eta), \\\\
\omega \wedge (a\eta + a'\eta') &= a(\omega \wedge \eta) + a'(\omega \wedge \eta').
\end{align*}
\]
- Associativity. \(\omega \wedge (\eta \wedge \xi) = (\omega \wedge \eta) \wedge \xi\).
- Anticommutativity. \(\omega \wedge \eta = (-1)^{kl}\,\eta \wedge \omega\).
- For covectors \(\omega^1, \dots, \omega^k \in V^*\) and vectors \(v_1, \dots, v_k \in V\),
\[
(\omega^1 \wedge \cdots \wedge \omega^k)(v_1, \dots, v_k) = \det\bigl(\omega^j(v_i)\bigr),
\]
the determinant of the \(k \times k\) matrix with \((i,j)\)-entry \(\omega^j(v_i)\).
Proof:
Bilinearity.
The tensor product is bilinear in its two factors, and the alternation operator
is linear, so the composite
\(\omega \wedge \eta = \tfrac{(k+l)!}{k!\,l!}\operatorname{Alt}(\omega \otimes \eta)\) is bilinear in \(\omega\) and \(\eta\).
Associativity.
By bilinearity, it suffices to verify associativity on basis elements,
since every alternating tensor is a linear combination of elementary tensors \(\varepsilon^I\).
For these, the concatenation rule \(\varepsilon^I \wedge \varepsilon^J = \varepsilon^{IJ}\) applies repeatedly:
\[
\begin{align*}
(\varepsilon^I \wedge \varepsilon^J) \wedge \varepsilon^K
&= \varepsilon^{IJ} \wedge \varepsilon^K \\\\
&= \varepsilon^{IJK} \\\\
&= \varepsilon^I \wedge \varepsilon^{JK} \\\\
&= \varepsilon^I \wedge (\varepsilon^J \wedge \varepsilon^K),
\end{align*}
\]
since concatenation of multi-indices is associative. Bilinearity extends the identity from basis elements
to all alternating tensors.
Anticommutativity.
Again by bilinearity it suffices to treat elementary tensors \(\varepsilon^I\) and \(\varepsilon^J\)
with \(I\) of length \(k\) and \(J\) of length \(l\).
The concatenation rule gives
\(\varepsilon^I \wedge \varepsilon^J = \varepsilon^{IJ}\) and \(\varepsilon^J \wedge \varepsilon^I = \varepsilon^{JI}\).
The multi-index \(JI\) is obtained from \(IJ\) by moving each of the \(l\) entries of \(J\) leftward past all \(k\) entries of \(I\),
a total of \(kl\) transpositions of adjacent entries. Each transposition of two indices changes the sign of the elementary tensor,
so \(\varepsilon^{JI} = (-1)^{kl}\,\varepsilon^{IJ}\), that is,
\[
\varepsilon^J \wedge \varepsilon^I = (-1)^{kl}\,\varepsilon^I \wedge \varepsilon^J.
\]
Rearranging gives the stated identity.
Part (4).
Each covector \(\omega^j\) is an alternating \(1\)-tensor, and iterating the definition of the wedge product across \(k\)
factors of degree \(1\) produces the coefficient \(k!\) in front of the alternation,
\[
\omega^1 \wedge \cdots \wedge \omega^k = k!\,\operatorname{Alt}(\omega^1 \otimes \cdots \otimes \omega^k).
\]
Expanding the alternation, the factor \(k!\) cancels the \(1/k!\) it carries, leaving
\[
(\omega^1 \wedge \cdots \wedge \omega^k)(v_1, \dots, v_k)
= \sum_{\sigma \in S_k} (\operatorname{sgn}\sigma)\,\omega^1(v_{\sigma(1)}) \cdots \omega^k(v_{\sigma(k)}).
\]
The right-hand side is exactly the Leibniz expansion of the determinant of the matrix whose \((i, j)\)-entry
is \(\omega^j(v_i)\): the term for \(\sigma\) contributes
\((\operatorname{sgn}\sigma)\) times the product \(\prod_j \omega^j(v_{\sigma(j)})\), and summing over \(\sigma\)
reproduces the determinant. Hence the value equals \(\det\bigl(\omega^j(v_i)\bigr)\).
Part (4) is the formula that makes wedge products of covectors computable: it identifies \(\omega^1 \wedge \cdots \wedge \omega^k\)
with a determinant of evaluations, recovering the elementary-tensor definition as the special case \(\omega^j = \varepsilon^{i_j}\).
Anticommutativity carries an immediate consequence worth isolating: a wedge of covectors in which any factor is repeated must vanish.
Repeated Factors and Odd Degree
Setting \(\omega = \eta\) of odd degree \(k\) in the anticommutativity rule gives
\(\omega \wedge \omega = (-1)^{k^2}\,\omega \wedge \omega = -\,\omega \wedge \omega\), since \(k^2\) is odd, forcing
\(\omega \wedge \omega = 0\). In particular every \(1\)-form squares to zero under the wedge:
\(\varepsilon^i \wedge \varepsilon^i = 0\), recovering the rule that an elementary tensor with a repeated index vanishes.
More generally, any wedge product containing two equal factors of odd degree is zero — the algebraic shadow of the fact that
a parallelepiped with two coincident edges is degenerate and bounds no volume.
The Exterior Algebra
Bilinearity and associativity make the wedge product more than an operation between fixed degrees:
assembling the alternating tensors of every degree into one space turns them into an algebra,
with the wedge as its multiplication. This is the exterior algebra, the natural home
of all the alternating tensors on \(V\) at once.
Definition: The Exterior Algebra
Let \(V\) be an \(n\)-dimensional vector space. The exterior algebra of \(V^*\) is the direct sum of
the spaces of alternating tensors of all degrees,
\[
\Lambda^*(V^*) = \bigoplus_{k=0}^{n} \Lambda^k(V^*),
\]
with the convention that \(\Lambda^0(V^*) = \mathbb{R}\) (scalars) and \(\Lambda^1(V^*) = V^*\) (covectors).
The wedge product extends to this sum by bilinearity, making \(\Lambda^*(V^*)\) an associative algebra
in which the product of a degree-\(k\) element and a degree-\(l\) element has degree \(k+l\).
An algebra with this degree-additive multiplication is called graded, and
because the wedge satisfies \(\omega \wedge \eta = (-1)^{kl}\,\eta \wedge \omega\), the exterior algebra is
graded anticommutative.
The sum terminates at \(k = n\) because \(\Lambda^k(V^*) = \{0\}\) for \(k \gt n\): no nonzero alternating tensor has
degree exceeding the dimension of \(V\). Summing the dimensions of the graded pieces gives the dimension of the whole algebra,
\[
\dim \Lambda^*(V^*) = \sum_{k=0}^{n} \dim \Lambda^k(V^*) = \sum_{k=0}^{n} \binom{n}{k} = 2^n,
\]
the last equality being the binomial expansion of \((1+1)^n\).
A basis for the entire algebra is the collection of all wedge products \(\varepsilon^{i_1} \wedge \cdots \wedge \varepsilon^{i_k}\)
over increasing multi-indices of every length from \(0\) to \(n\), the empty wedge being the scalar \(1 \in \Lambda^0(V^*)\).
The \(2^n\) such products correspond to the \(2^n\) subsets of \(\{1, \dots, n\}\).
One point of convention deserves mention, because two incompatible normalizations of the wedge product appear in the literature.
The definition used here carries the coefficient \((k+l)!/(k!\,l!)\), chosen so that wedges of elementary tensors concatenate
cleanly and the determinant formula of Part (4) holds with no extra factor. An alternative convention omits this coefficient,
defining the product directly as the alternation \(\operatorname{Alt}(\omega \otimes \eta)\). The two products differ only
by a degree-dependent scalar factor on each pair of degrees, so they generate the same subspaces and the same notion
of which forms are decomposable; the determinant normalization is adopted throughout this development so that the formulas
for differential forms match the classical change-of-variables and integration statements without correction terms.
Interior Multiplication
The wedge product raises degree. There is a companion operation that lowers it: feeding a fixed vector
into the first argument of an alternating tensor produces an alternating tensor of one degree less. This operation,
interior multiplication, pairs with the wedge product through an anti-derivation rule, and
it will reappear in the dynamics of forms as the contraction that turns a flow into an infinitesimal change of a form.
Definition: Interior Multiplication
Let \(v \in V\). The interior multiplication by \(v\) is the map
\(\iota_v : \Lambda^k(V^*) \to \Lambda^{k-1}(V^*)\) that inserts \(v\) into the first slot:
\[
(\iota_v \omega)(v_1, \dots, v_{k-1}) = \omega(v, v_1, \dots, v_{k-1}).
\]
It is also written \(v \lrcorner\, \omega\). On a \(0\)-tensor (a scalar) the operation is defined to be zero.
Because \(\omega\) is alternating, so is \(\iota_v \omega\): interchanging two of its arguments interchanges two arguments
of \(\omega\) among the slots after the first.
Interior multiplication is linear both in the form and in the vector. Two further properties characterize
how it interacts with the algebra structure: applying it twice gives zero, and it acts on a wedge product
as an anti-derivation.
Theorem: Properties of Interior Multiplication
Let \(v \in V\), let \(\omega \in \Lambda^k(V^*)\), and let \(\eta \in \Lambda^l(V^*)\).
- Applying interior multiplication twice annihilates: \(\iota_v(\iota_v \omega) = 0\), equivalently \(\iota_v \circ \iota_v = 0\).
- Interior multiplication is an anti-derivation with respect to the wedge product:
\[
\iota_v(\omega \wedge \eta) = (\iota_v \omega) \wedge \eta + (-1)^k\,\omega \wedge (\iota_v \eta).
\]
Proof:
Part (1).
Evaluating \(\iota_v(\iota_v \omega)\) on \(k-2\) vectors \(v_1, \dots, v_{k-2}\),
\[
\begin{align*}
\bigl(\iota_v(\iota_v \omega)\bigr)(v_1, \dots, v_{k-2})
&= (\iota_v \omega)(v, v_1, \dots, v_{k-2}) \\\\
&= \omega(v, v, v_1, \dots, v_{k-2}).
\end{align*}
\]
The argument \(v\) appears twice, and an alternating tensor vanishes whenever two of its arguments coincide.
Hence the result is zero for every input, so \(\iota_v \circ \iota_v = 0\).
Part (2).
Both sides are linear in \(\omega\) and in \(\eta\), so by bilinearity of the wedge product it suffices
to verify the identity on elementary tensors, and indeed on wedges of covectors, since every elementary tensor is such a wedge.
We first record the action of \(\iota_v\) on a single covector and on a wedge of covectors.
For a covector \(\alpha \in V^* = \Lambda^1(V^*)\), the definition gives \(\iota_v \alpha = \alpha(v)\), a scalar.
For a wedge \(\alpha^1 \wedge \cdots \wedge \alpha^k\) of covectors, the determinant formula expresses its value
on \((v, v_1, \dots, v_{k-1})\) as a determinant whose first column is \((\alpha^1(v), \dots, \alpha^k(v))^\top\).
Expanding that determinant along the first column,
\[
\iota_v(\alpha^1 \wedge \cdots \wedge \alpha^k)
= \sum_{i=1}^{k} (-1)^{i-1}\,\alpha^i(v)\;\alpha^1 \wedge \cdots \wedge \widehat{\alpha^i} \wedge \cdots \wedge \alpha^k,
\]
where the hat denotes omission of the \(i\)th factor.
This is the cofactor expansion read as an identity of \((k-1)\)-tensors.
Now write \(\omega = \alpha^1 \wedge \cdots \wedge \alpha^k\) and \(\eta = \beta^1 \wedge \cdots \wedge \beta^l\), so that
\(\omega \wedge \eta\) is the wedge of all \(k+l\) covectors in order. Applying the expansion above to \(\omega \wedge \eta\),
the terms in which the omitted factor lies among the \(\alpha\)'s reassemble into \((\iota_v \omega) \wedge \eta\), each carrying
its original sign \((-1)^{i-1}\). The terms in which the omitted factor lies among the \(\beta\)'s carry sign \((-1)^{k + j - 1}\),
because the omitted \(\beta^j\) sits in position \(k + j\); factoring out \((-1)^k\) leaves \((-1)^{j-1}\), and these terms
reassemble into \((-1)^k\,\omega \wedge (\iota_v \eta)\). Summing the two groups gives
\[
\iota_v(\omega \wedge \eta) = (\iota_v \omega) \wedge \eta + (-1)^k\,\omega \wedge (\iota_v \eta),
\]
as claimed. Linearity extends the identity from wedges of covectors to all alternating tensors.
The sign \((-1)^k\) in the anti-derivation rule records the cost of moving the inserted vector past the \(k\) arguments of \(\omega\)
to reach the arguments of \(\eta\) — the same bookkeeping of transpositions that governs anticommutativity.
Together, raising degree by wedging and lowering it by interior multiplication equip the exterior algebra
with the two operations from which the calculus of differential forms is built. When the fixed vector is replaced
by a vector field and the alternating tensors by smoothly varying fields on a manifold, interior multiplication
becomes the contraction appearing in the identity that links it to the exterior derivative and the Lie derivative.