We ask for $K$ not to have characteristic $2$, for this section.

Remember that $V^{\otimes k}$ is the space of homogeneous tensors of degree $k$. This is spanned by decomposable tensors $$v_1 \otimes v_2 \otimes \dots \otimes v_k,\quad v_i \in V$$
Def: The antisymmetrisation or skew-symmetrisation of decomposable tensors is defined by $$\mathcal A(v_1 \otimes \dots \otimes v_k) := \sum_{\sigma \in S_k} \text{sgn}(\sigma) v_{\sigma(1)} \otimes \dots \otimes v_{\sigma(k)}$$where the sum is taken over the symmetric group and, when the $k!\ne 0$: $$\text{Alt}(v_1 \otimes \dots\otimes v_k) := \frac1{k!} \sum_{\sigma \in S_k} \text{sgn}(\sigma) v_{\sigma(1)} \otimes \dots \otimes v_{\sigma(k)}$$
Def: In the language of tensors, a $p$-vector is an alternating contravariant tensor of order $p.$ A $p$-vector is denoted by $A_{[p]}$ and characterised by $$A_{[p]} = \text{Alt}(A_{[p]}).$$The brackets here are used to indicate the alternation of the $p$ indices set. Thus, given $A_p\in V^{\otimes p}$, $\text{Alt}(A_p)$ is a $p$-vector. The symbols ${\bigwedge }_p(V)$ and ${\bigwedge }^p(V)$ respectively denote the space space of $p$-vectors and $p$-covectors.

Prop: Let $p\in \mathbb{N}$, then $$\dim {\textstyle \bigwedge}^{!p}(V) = \dim {\textstyle \bigwedge}p(V) = {n \choose p}.$$
Cor: Let $p\in \mathbb{N}$, then $$\dim {\bigwedge}p(V) = \dim {\bigwedge}(V).$$
Def: Let $A_{[p]}\in {\bigwedge }_p(V)$ be a $p$-vector and let $B_{[q]}\in {\bigwedge }_q(V)$ be a $q$ vector. The exterior product $\wedge :{\bigwedge }_p(V)\times {\bigwedge }_q(V)\to {\bigwedge }_{p+q}(V)$ is defined by $$A \wedge B_{[q]} := \text{Alt}(A_{[p]} \otimes B_{[q]}).$$
Prop: Let $A_{[p]}\in \underset{p}{\bigwedge }(V)$, $B_{[q]}\in {\bigwedge }_q(V)$ and $C_{[r]}\in {\bigwedge }_r(V)$, then $$(A_{[p]} \wedge B_{[q]}) \wedge C_{[r]} = A_{[p]} \wedge (B_{[q]} \wedge C_{[r]}),$$meaning that $\wedge$ is associative.

Def: For $p\in \mathbb{N}$, we can define a left action and a right action from $S_p$ on ${\bigwedge }_p(V)$.
Let $\sigma \in S_p$, and $A_{[p]}\in {\bigwedge }_p(V)$:

• $\sigma (v_1\otimes \cdots \otimes v_p)=\text{sgn}(\sigma )v_{{\sigma }^{-1}(1)}\otimes \dots v_{{\sigma }^{-1}(p)}.$
• $(v_1\otimes \cdots \otimes v_p{)}^{\sigma }=\text{sgn}(\sigma )v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (n)}.$

Obs: We get another characterisation of ${\bigwedge }_p(V)$, it is a subspace of $V^{\otimes p}$ such that for all $A_{[p]}$ either $\sigma (A_{[p]})=A_{[p]}$ or $(A_{[p]}{)}^{\sigma }=A_{[p]}$ for all $\sigma \in S_p$.

Obs: The definition of the exterior product is equivalent to $$A_{[p]} \wedge B_{[q]} := \frac{p!q!}{(p+q)!} \sum_{\sigma\in S_{p, q}} \sigma(A_{[p]} \otimes B_{[q]}), $$where $S_{p,q}$ denotes the subset of $S_{p+q}$ containing all $\sigma$ such that $\sigma (i)<\sigma (i+1)$ if $0<i<p$, or $p<i<p+q$.

Prop: From this equation, it follows that $u\wedge v=-v\wedge u$, and $v\wedge v=0$.

Cor: $A_{[p]}\wedge B_{[q]}=(-1{)}^{pq}{B}_{[q]}\wedge A_{[p]}$.

Def: A $p$-vector that can be written as the exterior product of a $p$ number of $1$-vectors is called a simple $p$-vector.

Bases

Suppose that $V$ is a finite dimensional $K$-vector space, and let $\mathfrak{B}=\{e_1,\dots ,e_n\}$ be a basis for $V$. We can construct the bases for ${\bigwedge }_p(V)$. First we need a shortcut in notation, let $\underset{i\in I}{\bigwedge }{v}_i$ is the exterior product of all the elements $\{v_i\mid i\in I\}$ where $I\subseteq \{1,\dots ,n\}$ in the order inherited by $I$. With this in mind, we can define the basis of ${\bigwedge }_p(V)$ as $$\mathfrak B_p := \left{\left.\bigwedge_{i \in I} e_i; \right\rvert ; I \in [{1, \dots, n}]^p\right}$$A consequence of this basis, if $A_{[2]}$, then $$A_{[2]} = \frac12 \sum_{i, j = 1}^n A^{ij} e_i \wedge e_j = \sum_{1 \le i < j \le n} A^{ij} e_i \wedge e_j.$$Notice in the first expression the presence of the factor of $1/2$, which is absent in the second expression. Since, in the sum $\sum _{ij}$, we consider all the values for the indices $i,j\in \{1,\dots ,n\}$, as $A^{ij}=-A^{ji}$, and as $e_i\wedge e_j=-e_j\wedge e_i$, we are indeed counting the same term twice.

We generalise this for ${\bigwedge }_p(V)$. An arbitrary element $A_{[p]}\in {\bigwedge }_p(V)$ can be written as $$
\begin{align*}
A_{[p]} &:= \frac{1}{p!} \sum_{\mu_1, \dots, \mu_p = 1}^n A^{\mu_1, \dots, \mu_p} e_{\mu_1} \wedge \dots \wedge e_{\mu_p} \
&= \sum_{1 \le \mu_1 <\dots < \mu_p \le n} A^{\mu_1, \dots, \mu_p} e_{\mu_1} \wedge \dots \wedge e_{\mu_p},
\end{align*}$$where, in the first case, we consider the sum over all posible values for the indices ${\mu }_i\in \{1,\dots ,n\}$ for all $i\in \{1,\dots ,p\}$, and, in the second case, we consider the sum over the indices with the restriction that ${\mu }_1<{\mu }_2<\cdots <{\mu }_p$.

Convention: In order to apply the sum convention to $p$-vectors as efficiently as possible, we impose the same restriction on the sum convention with respect to the indices, namely, that $$A^{\mu_1, \dots, \mu_p} e_{\mu_1} \wedge \dots \wedge e_{\mu_p} := \sum_{1 \le \mu_1 <\dots < \mu_p \le n} A^{\mu_1, \dots, \mu_p} e_{\mu_1} \wedge \dots \wedge e_{\mu_p}.$$Thus, we avoid using expressions with the factor $p!$.

An ordered $k$-tuple $I=(i_1,\dots ,i_k)$ of a positive integers is called a multi-index of length $k.$ If $I$ is such a multi-index and $\sigma \in S_k$ is a permutation we write $I_{\sigma }$ for the multi-index$$I = (i_{\sigma(1)},\dots, i_{\sigma(k)}).$$
We see that $I_{\sigma \tau }=(I_{\sigma }{)}_{\tau }$ for $\sigma ,\tau \in S_k$. We can extend the Kronecker delta notation in the following way. If $I$ and $J$ are multiindices of length $k$, we define $$\delta^I_J := \begin{cases}
\text{sgn }\sigma& \text{if neither $I$ nor $J$ has repeadted index and $J=I_{\sigma }$ for some $\sigma \in S_k$ }\
0 & \text{If $I$ or $J$ has repated index or $J$ is not a permutation of $I$}.
\end

Obs: If $J=I_{\sigma }$, for some multi-indices, then $e_J=(\text{sgn }\sigma )e_I$.

The significance of this that they provide a more convenient basis for $\underset{p}{\bigwedge }(V)$. We see that the set of $e_I$ is not linearly independent, because some of them are zero and the ones corresponding to different permutations of the same multi-index are constant multiples of each other. A multi-index $I=(i_1,\dots ,i_k)$ is said to be increasing if $i_1<\cdots <i_k$. This means that using Einstein summation convention we can write a $p$-vector as $$A^I e_I := \sum_{I}' A_I e_I = \sum_{{I: 1 \le \mu_1 <\dots < \mu_p \le n}} A^{\mu_1, \dots, \mu_p} e_{\mu_1} \wedge \dots \wedge e_{\mu_p}$$and the set $\{e_I\mid I\text{ is an increasing }p\text{multi-index}\}$ is a basis for $\underset{p}{\bigwedge }(V)$.

Def: Since ${\bigwedge }_n(V)$ is a one dimensional $K$-vector space, then we call the elements of ${\bigwedge }_n(V)$ are called $n$-vectors, and another usual denomination for $n$-vectors is psuedoscalars.

Def: The image $\text{Alt}[T(V)]$ is always the alternating tensor graded subspace, it is not algebra yet since we don't have a product, denoted as $A(V)$.

Obs: We have that the kernel of ${\mathcal{A}}^{(r)}$ is precisely $I^{(r)}$, the homogeneous subset of the ideal $I$, the kernel of $\mathcal{A}$ is $I$.

When $K$ has characteristic $0$, there is a canonical isomorphism $$A(V) \cong {\textstyle \bigwedge}(V).$$With these isomorphisms, we see that exterior algebra of a vector space $V$ over $K$ can be canonically identifies with the vector subspace of $T(V)$ that consists of antisymmetric tensors. With this isomorphism in mind, we don't need to care about ${\bigwedge }_p(V)$ as a subspace of $A(V)$ or $\bigwedge (V)$.

Cor: $\dim \bigwedge (V)=2^n$.

Def: We denote by $⟨\cdot {⟩}_p:\bigwedge (V)\to {\bigwedge }_p(V)$ as a projection from $\bigwedge (V)$ to a space of $p$-vectors $⟨A{⟩}_p:=A_{[p]}$.

Def: We can define grade involution on ${\bigwedge }_p(V)$: $$#(A_{[p]}) = \widehat A_{[p] } := (-1)^p A_{[p]}. $$For the reversion, it follows that $$\widetilde{(v_1\wedge \dots \wedge v_p)} := v_p \wedge \dots \wedge v_1, $$this implies that ${\tilde{A}}_{[p]}=(-1{)}^{p(p-1)/2}{A}_{[p]}$. The conjugation of the two operations $$\overline A_{[p]} := \widetilde{\widehat A}{[p]} = \widehat{\widetilde A}.$$

Contraction, or Interior Product

Let $A_{[p]}$ be a $p$-vector and let $\alpha$ be a covector.

Def: The left contraction of a $p$-vector $A_{[p]}$ by a covector $\alpha$, denoted from this point on by $\alpha \rfloor$, is defined as $$(\alpha \rfloor A_{[p]})(\alpha_1, \dots, \alpha_{p-1}) := p A_{[p]}(\alpha, \alpha_1, \alpha_2, \dots, \alpha_{p-1}).$$where ${\alpha }_1,\dots ,{\alpha }_{p-1}$ are arbitrary covectors. Taking $A_{[p]}=v_1\wedge \cdots \wedge v_p$ on the right hand side of the definition means that $$(v_1\wedge \dots \wedge v_p)(\alpha, \alpha_1, \dots, \alpha_{p-1}) := \frac1{p!} \sum_{\sigma \in S_p} \text{sgn}(\sigma) \alpha(v_{\sigma(1)}) \alpha_1(v_{\sigma(2)})\dots \alpha_{p-1}(v_{\sigma(p)}).$$
This definition shows that $\alpha \rfloor A_{[p]}$ is a $(p-1)$-vector.

Lemma: Let $u,v\in V$ and $\alpha \in V^{\prime }$, then $$\alpha \rfloor (v \wedge u) = (\alpha \rfloor v)u - v(\alpha \rfloor u) = \alpha(v) u - v \alpha (u).$$
Prop: Let $A,B\in \bigwedge (V)$ and $\alpha \in V^{\prime }$, then $$(\alpha \rfloor A \wedge B) = (\alpha \rfloor A) \wedge B +\widehat A \wedge (\alpha \rfloor B)

Obs: Given the definition above, let $v\in V$ and $\alpha \in V^{\prime }$, then $\alpha \rfloor v=\alpha (v)=v\lfloor \alpha$.

Prop: Let $A,B\in \bigwedge (V)$, and $\alpha \in V^{\prime }$, then $$(A \wedge B) \lfloor \alpha = A \wedge (B \lfloor \alpha ) + (A \lfloor \alpha ) \wedge\widehat B.$$
Prop: Let $A\in \bigwedge (V)$, and $\alpha \in V^{\prime }$, then $$\alpha \rfloor A = - \widehat A \lfloor \alpha.$$
We can generalise further the left and right contractions to arbitrary covectors.

Def: Given ${\alpha }_1\wedge \cdots \wedge {\alpha }_p$, we can define $$(\alpha_1 \wedge \alpha_2 \wedge \dots \wedge \alpha_p) \rfloor := \alpha_1 \rfloor \alpha_2\rfloor\cdots \alpha_p\rfloor,$$and $$\lfloor (\alpha_1 \wedge \alpha_2 \wedge \dots \wedge \alpha_p) := \lfloor\alpha_1 \lfloor\alpha_2 \cdots \lfloor \alpha_p.$$
Obs: Using this definition of the left contraction of a $2$-covector on $2$-vector we get that $$(\alpha \wedge \beta)\rfloor (u \wedge v) = \alpha(v) \beta(u)- \alpha (u) \beta(v) = 2(\alpha\wedge \beta)(u \wedge v). $$This can generalised further. Let ${\mathrm{\Phi }}^{[p]}$ be a $p$-covector and $A_{[p]}$ a $p$-vector, then $$\Psi^{[p]} (A_{[p]}) = \frac1{p!} \Psi^{[p]}\rfloor A_{[p]}. $$
Prop: Let ${\mathrm{\Psi }}^{[p]}$ be a $p$-covector and $A_{[q]}$ a $q$-vector. If $p\le q$, then $$\Psi^{{[p]}}\rfloor A_{[q]} = (-1)^{p(q-1)} A_{[q]} \lfloor \Psi^{[p]}.$$Similarly, if $q\le p$, then $$A_{[q]} \rfloor \Psi^{[p]} = (-1)^{q(p-1)} \Psi^{[p]}\lfloor A_{[q]}.$$Lastly, $$\widetilde{\Psi^{[p]}\rfloor A_{[q]}} = A_{[q]}\lfloor \Psi^{[p]}. $$

Obs: Let us note the equality $$\Psi^{[p]}\rfloor A_{[p]} = A_{[p]} \lfloor\Psi^{[p]} = A_{[p]} \rfloor \Psi^{[p]} = \Psi^{[p]} \lfloor A_{[p]}.$$

Orientation

In this section we are considering that $K=\mathbb{R}$.

There is no criteria that would induce us to prefer the choice of some specific $n$-vector. Ia basis $\mathfrak{B}:=\{e_1,\dots ,e_n\}$ of $V$ is taken into account, it is natural to choose the $n$-vector given by the exterior product of the basis vector $\mathfrak{B}$. No matter the order that such exterior products are taken $e_{\sigma (1)}\wedge \cdots \wedge e_{\sigma (n)}$ for every $\sigma \in S_n$, the result is $\pm e_1\wedge \cdots \wedge e_n$.

Obs: Let $\{v_1,\dots ,v_n\}$ be $n$-linearly independent vectors, then $v_1\wedge \cdots \wedge v_n=J{e}_1\wedge \cdots \wedge e_n.$ where $J>0$ or $J<0$.

Def: $\mathrm{\Omega }$ be a $n$-vector, then we can define two equivalence classes $[\mathrm{\Omega }]$ and $[-\mathrm{\Omega }]$, as $$\begin{align*}
[\Omega] &:= \left{\left. A \in { \bigwedge}_n (V);\right\rvert; \exists J >0(A =J\Omega) \right} \
[-\Omega] &:= \left{\left. A \in {\bigwedge}_n (V);\right\rvert; \exists J <0(A =J\Omega) \right}
\end{align*}.$$These two equivalence classes compromise the vector space orientation.

There exists two possible orientations for a vector space, completely determined once ab $n$-vector has been selected. The choice of an $n$-vector ${\mathrm{\Omega }}_V$ in an equivalence class defines a pair constituted by a vector space $V$ endowed with an $n$-vector ${\mathrm{\Omega }}_V$.

We can define an orientation for the dual space $V^{\prime }$ via de $n$-covector ${\mathrm{\Omega }}_{V^{\ast }}$. The question to be posed is whether the orientation of $V^{\prime }$ can be related to the orientation of $V$.

Duality

If we consider the set of all alternating $k$-linear forms, meaning the set of alternating $k$-linear maps from $V$ to the base field $K$. We see that the set of all alternating $k$-linear forms is a vector space.

By the universal property the exterior powers, the space of alternating forms of degree $k$ on $V$ is naturally isomorphic with the dual space $({\bigwedge }_k(V){)}^{\ast }$. Lastly, if $V$ is finite dimensional, then the latter is naturally isomorphic to ${\bigwedge }_k(V^{\ast })$. Using the notation, from Exterior Algebra of Multicovectors, then we get that ${\bigwedge }^k(V)=A_k(V)\cong {\bigwedge }_k(V^{\ast })$.