Exterior Algebra as a quotient is or is not compatible with Exterior Algebra as a vector subspace space with $\det$ convention of wedge product?

Question

Ok, so I asked a similar question here, but after the first response it quickly devolved into an unfocused mess. I now believe I can adequately formulate my confusion.

Let $V$ be $\mathbb{K}=\mathbb{C},\mathbb{R}$ linear vector space of dimension $n$, and $V^*$ it's dual. I am going to work with the dual space $V^*$ for convenience, but since $(V^*)^*$ is canonically isomorphic to $V$ everything I do here can be done in with $V$ by just replacing vectors with covectors and vice versa. We denote the exterior algebra of $V^*$ by: $$\Lambda(V^*)=\bigoplus_{k=0}^n\Lambda^k(V^*)$$ where each $\Lambda^k(V)$ is the subspace of multilinear maps $V^k\rightarrow \mathbb{K}$ which are totally antisymmetric. The antisymetrization map: $$ \begin{align} \text{Alt}:(V^*)^{\otimes k}&\longrightarrow \Lambda^k(V^*) \end{align} $$ is defined by it's action on a set of vectors $v_1,\dots,v_k\in V$: $$\text{Alt}(\omega)(v_1,\dots,v_k)=\frac{1}{k!}\sum_{\sigma\in S_k}\text{sgn}(\sigma)\omega(v_{\sigma(1)},\dots,v_{\sigma(k)})$$ We define a wedge product on $\Lambda(V^*)$ by: $$\omega\wedge \eta=\frac{(k+l)!}{k!l!}\text{Alt}(\omega\otimes \eta)$$ where $\omega\in \Lambda^k(V^*)$ and $\eta\in \Lambda^k(V^*)$. This product is associative, and turns $\Lambda(V^*)$ into a graded, associative $\mathbb{K}$ algebra with unit element $1\in\mathbb{K}$. Furthermore, one can show that for a set of $k$ covectors $\omega^1,\cdots, \omega^k\in V^*$, the wedge product satisfies: $$\omega^1\wedge \cdots \wedge \omega^k=\sum_{\sigma\in S_k}\text{sgn}(\sigma)\omega^{\sigma(1)}\otimes\cdots\otimes\omega^{\sigma(k)}$$ I did this proof in edit 3 of my other question, and I would not be surprised if I messed something up. However, even if something is wrong with the proof, I strongly suspect that the above still holds as it seems equivalent to $\textbf{Proposition 14.11 part d)}$ in Lee's smooth manifolds. This property is also why it is called the determinant convention, as if $k=n$, and $\omega^i=e^i$, where $\{e^i\}$ is the basis for $V^*$ dual to the basis $\{e_i\}$ for $V$, then the above product is the determinant of $n$ vectors.

Now, let $$T(V^*)=\bigoplus_{n=0}^\infty (V^*)^{\otimes n}$$ and let $I$ be the ideal generated by: $$\{\omega\otimes \omega |\omega\in V^*\}$$ Then since $I$ is a graded ideal, as it is generated by homogenous elements, and $T(V^*)$ is a graded associative $\mathbb{K}$ algebra with unit element $1$, it follows that with $I_k=I\cap(V^*)^{\otimes k}$: $$T(V^*)/I=\bigoplus_{k=0}^\infty T(V^*)_k/I_k$$ is a graded associative $\mathbb{K}$ algebra with unit element $1$. Furthermore, it is not hard to show that for $k>n$, $T(V^*)_k/I_k=\{0\}$, and that for $k=0,1$, $T(V^*)_0/I_0=\mathbb{K}$, and $T(V^*)_1/I_1=V^*$. For each $2\leq k\leq n$, we then define the linear map: $$\phi^k:(V^*)^{\otimes k}\longrightarrow \Lambda^k(V^*)$$ which on simple tensors is given by: $$\omega^1\otimes\cdots \otimes\omega^k\longrightarrow \omega^1\wedge \cdots \wedge \omega^k$$. The map is easily seen to be surjective, and since any $i\in I_k$ can be written as the sum: $$i=\sum_i\sum_{j+l=k-2}a_j\otimes(\omega_i\otimes \omega_i)\otimes b_l$$ where each $a_j\in V^{\otimes j}$ and each $b_l\in V^{\otimes l}$, we have that $i\in \ker\phi^k$, so $I_k\subset \ker \phi$, and by the universal property of quotient vector spaces we have that $\phi_k$ descends to a surjective linear map $\psi_k:T(V^*)/I_k\rightarrow \Lambda^k(V^*)$. After deriving some properties of multiplication in $T(V)/I$, which are pretty much the same as the properties of the wedge product, it is also not difficult to see that the set: $$B=\{[e^{i_1}\otimes\cdots \otimes e^{i_k}]:i_1<\cdots<i_k\}$$ spans $T(V)_k/I_k$. It is linearly independent since for any linear combination combination: $$\sum_{i_1<\cdots<i_k}a_{i_1\cdots i_k}[e^{i_1}\otimes\cdots \otimes e^{i_k}]=0$$ we have that under the map$\psi^k$: $$\sum_{i_1<\cdots <i_k}a_{i_1\cdots i_k}e^{i_1}\wedge \cdots \wedge e^{i_k}=0$$ Hence, if $j_1\cdots j_k$ is any other ordered multi index, we have that: $$ \begin{align} \sum_{i_1<\cdots <i_k}a_{i_1\cdots i_k}e^{i_1}\wedge \cdots \wedge e^{i_k}(e_{j_1},\cdots e_{j_k})=&\sum_{i_1\cdots i_k}a_{i_1\cdots i_k}\delta^{i_1\cdots i_k}_{j_1\cdots j_k}\\ =&a_{j_1\cdots j_k}\\ =&0 \end{align} $$ So each coefficient in the sum is zero, and $B$ is linearly independent. Since the size of $B$ is clearly $n$ choose $k$, we have that $\dim T(V^*)_k/I_k=\dim \Lambda^k(V^*)$, so since $\psi^k$ is surjective, it follows by rank nullity that for each $k$: $$T(V^*)_k/I_k\cong \Lambda^k(V^*)$$

My question is then this, why can't we use these maps to create an algebra isomorphism between $T(V^*)/I$ and $\Lambda(V^*)$? From the reply in my previous post, it seems to suggest that this is impossible with the wedge product defined as I have it. However, I believe we can. Every element $a\in T(V^*)/I$ can be written as the sum: $$a=\sum_{i=0}^na_i$$ where each $a_i\in T(V^*)_i/T_i$, hence we use the $\psi^k$'s to define the following map: $$\psi:\sum_{i=0}^na_i\longrightarrow \sum_{i=0}^n\psi^i(a_i)$$ It is a linear bijection, so as vector spaces we have an isomorphism. By linearity, it suffices to check that one simple products we have that: $$\psi([\omega^1\otimes\cdots \otimes \omega^k]\cdot [\omega^{k+1}\otimes \cdots \otimes \omega^l])=\psi([\omega^1\otimes \cdots \otimes \omega^{k+1}])\wedge \psi([\omega^{k+1}\otimes \cdots \otimes \omega^{k+l}])$$ By the induced multiplicative structure on $T(V)/I$ we have that: $$[\omega^1\otimes\cdots \otimes \omega^k]\cdot [\omega^{k+1}\otimes \cdots \otimes \omega^l]=[\omega^1\otimes \cdots\omega^{k}\otimes\omega^{k+1} \otimes \omega^{k+l}]$$ hence: $$ \begin{align} \psi([\omega^1\otimes\cdots \otimes \omega^k]\cdot [\omega^{k+1}\otimes \cdots \otimes \omega^l])=&\psi^{k+l}([\omega^1\otimes \cdots\omega^{k}\otimes\omega^{k+1} \otimes \omega^{k+l}])\\ =&\omega^1\wedge \cdots \wedge \omega^k\wedge \omega^{k+1}\wedge \cdots \wedge \omega^{k+l} \end{align}$$ While: $$ \begin{align} \psi([\omega^1\otimes \cdots \otimes \omega^k])\wedge \psi([\omega^{k+1}\otimes \cdots \otimes \omega^{l}])=&\psi^k([\omega^1\otimes \cdots \otimes \omega^k])\wedge \psi^l([\omega^{k+1}\otimes \cdots \otimes \omega^{k+l}])\\ =&(\omega^1\wedge \cdots \wedge \omega^{k})\wedge (\omega^{k+1}\wedge \cdots \wedge \omega^{k+l})\\ =&\omega^1\wedge \cdots \wedge \omega^k\wedge \omega^{k+1}\wedge \cdots \wedge \omega^{k+l} \end{align}$$ so $\psi$ is respects multiplication in both algebras. It is clear that $\psi(1)=1$, as $\psi^0$ is the identity map, so $\psi$ is a $\mathbb{K}$-algebra homomorphism, and thus an isomorphism.

Am I missing something here? It seems that once we get a vector space isomorphism on each grade everything falls into place.

Answer 1

Although, as the OP points out, since we work with finite-dimensional vector spaces, there is no significant difference in working with the tensor algebra $\mathcal T(V) = \bigoplus_{k \geq 0} T^k(V)$ of $V$ or with the tensor algebra $\mathcal T(V^*)$ of $V^*$, I think the best way to understand the relationship between the two constructions is to view them as dual to each other, as I explain below: We will also describe the two constructions in a manner that works uniformly for any field $\mathsf k$. In particular, the constructions below work in all characteristics, including the case $\text{char}(\mathsf k)=2$.

Notation: Let $\mathsf k$ be an arbitrary field. For positive integers $n\leq m$ we set $[n,m] =\{n,n+1,\ldots,m\}$. For $V$ a $\mathsf k$-vector space and $k$ a non-negative integer $k$, we write $T^k(V)$ for the $k$-fold tensor product of $V$ with itself. For positive integers $n\leq m$, we will write $[n,m]:= \{n,n+1,2,\ldots,m\}$.

There is a natural $2k$-multilinear map $ V^k \times (V^*)^k \to \mathsf k$ given by $$ (v_1,\ldots,v_k,f_1,\ldots,f_k) \mapsto \prod_{j=1}^k f_j(v_j). $$ The universal property of tensor products shows that this induces a bilinear pairing $\langle -,-\rangle\colon T^k(V)\times T^k(V^*)\to \mathsf k$ and hence yields a map $\theta\colon T^k(V^*)\to T^k(V)^*$. If $V$ is finite-dimensional then $\theta$ is an isomorphism (as can be checked using dual bases).

The exterior algebra: Let $V$ be a finite-dimensional vector space and let $R = \text{span}\{v\otimes v: v \in V\}\subseteq V\otimes V$. The exterior algebra $\bigwedge(V)$ is defined to be the quotient of the tensor algebra $T(V)= \bigoplus_{k\geq 0} T^k(V)$ by the two-sided ideal $J=J(V)$ generated by $R$. If we set $J_k = T^k(V)\cap J$, then $$ \bigwedge(V) = \bigoplus_{k \geq 0}\bigwedge\!{^k(}V), \quad \text{where } \bigwedge\!{^k(}V) = T^k(V)/J_k. $$ As a quotient of the tensor algebra by a graded ideal, it is clearly a graded associative algebra.

Alternating forms: Via $\theta$, we may view $T^k(V^*)$ as the dual space of $T^k(V)$, or equivalently, as the space of all $k$-multilinear map $V^k \to \mathsf k$. Let $\Lambda(V^*) = \bigoplus_{k\geq 0} \Lambda^k(V^*)$, the space of alternating multilinear forms on $V$, where $$ \Lambda^k(V^*) = \{t \in T^k(V^*): t(v_1,\ldots,v_k)=0 \text{ if } \exists i \in [1,k], v_i=v_{i+1}\}. $$

Note that this definition of an alternating form can be restated as $\Lambda^k(V^*) = J_k^0$, the annihilator of $J_k$. Since $J_k^0$ is the image of $\bigwedge\!\!{^k(}V)^*$ under $q^t$, the transpose of the quotient map $q\colon T^k(V)\to \bigwedge\!\!{^k(}V)$, it follows $\Lambda^k(V^*)$ is naturally identified with $\bigwedge\!{^k(}V)^*$. We can therefore think of the space of alternating forms $\Lambda(V^*) = \bigoplus_{k \geq 0} \Lambda^k(V^*)$ as the "graded dual" of the exterior algebra.

The symmetric group action: For any vector space $U$, the symmetric group $S_k$ acts naturally on $T^k(U)$: if $t=u_1\otimes \ldots \otimes u_k$ then $\sigma(t) = u_{\sigma(1)}\otimes \ldots\otimes u_{\sigma(k)}$. Note that the pairing $\langle-,-\rangle$ is invariant with respect to this action, that is, $$ \tag{$\dagger$} \langle \sigma(s),\sigma(t) \rangle = \langle s,t \rangle, \quad \forall s \in T^k(V), t \in T^k(V^*), \sigma \in S_k. $$ We now analyse $\Lambda^k(V^*)$ using the symmetric group: At first sight, this may seem surprising since it is not immediately clear whether or not the action of $S_k$ even preserves the subspace $\Lambda^k(V^*)$!

To see that this is indeed the case, if $t\in \Lambda^k(V^*)$, $i \in \{1,\ldots,k-1\}$ and we have vectors $v_1,\ldots,v_k \in V$, writing $t_i(w_1,w_2)$ for $t(v_1,\ldots,v_{i-1},w_1,w_2,v_{i+2},\ldots,v_k)$ we have $$ \begin{split} 0&=t_i(v_i+v_{i+1},v_i+v_{i+1})\\ &= t_i(v_i,v_i) + t_i(v_i,v_{i+1})+t_i(v_{i+1},v_i)+ t_i(v_{i+1},v_{i+1})\\ &= 0+ t_i(v_i,v_{i+1}) + t_i(v_{i+1},v_i)+0. \end{split} $$ so that $t_i(v_i,v_{i+1}) = -t_i(v_{i+1},v_i)$. It follows that if $\sigma_i\in S_k$ is the transposition $(i,i+1)$, then for any $t \in \Lambda^k(V^*)$ we have $\sigma_i(t) = -t$. Thus we see that $\Lambda^k(V^*)$ is preserved by the $\sigma_i$ and hence, since $S_k=\langle \sigma_i: 1\leq i \leq k-1\rangle$, it follows that $\Lambda^k(V^*)$ is preserved by all of $S_k$. Indeed if $\varepsilon\colon S_k \to \{\pm 1\}$ is the sign character then since $\varepsilon(\sigma_i)=-1$ for all $i \in [1,k-1]$ it follows that if $T \in \Lambda^k(V^*)$ then $\sigma(T)=\varepsilon(\sigma).T$.

Equivalently, if we set $T^k_{\varepsilon}(V^*) = \{T \in T^k(V^*): \sigma(T) = \varepsilon(\sigma).T\}$, we have $\Lambda^k(V^*) \subseteq T_{\varepsilon}^k(V^*)$. Note that this implies that if $T \in \Lambda^k(V^*)$ and $(v_1,\ldots,v_k)$ is any $k$-tuple of vectors such that there is a pair $(i,j)$ with $1\leq i <j\leq k$ with $v_i = v_j$ then $T(v_1,\ldots,v_k)=0$. Indeed this is true by definition if $j=i+1$, that is, if $v_i =v_{i+1}$, but since the value of $T$ changes by at most a sign when we permute the $v_i$s it follows that $T$ must vanish whenever the vectors in the $k$-tuple $(v_1,\ldots,v_k)$ are not pairwise distinct.

Bases of tensor spaces If $\{e_1,\ldots,e_n\}$ is a basis of $V$ and $\{x_1,\ldots,x_n\}$ the corresponding dual basis of $V^*$, then for any $v \in V$ we have $v = \sum_{i=1}^n x_i(v)e_i$. Thus if we let $$ [1,n]^k = \{I=(i_1,\ldots,i_k): 1\leq i_j\leq n, \forall j, 1 \leq j \leq k\} $$ and for $I \in [1,n]^k$, $\vec{e}_I = (e_{i_1},\ldots,e_{i_k})$ then if $T \in T^k(V^*)$ we see that $$ T(v_1,\ldots,v_k) = \sum_{I \in [1,n]^k}T(\vec{e}_I)x_{i_1}(v_1)\ldots x_{i_k}(v_k). $$ so that if we set $x_I= x_{i_1}\otimes \ldots \otimes x_{i_k} \in T^k(V^*)$ it follows that $\{x_I: I \in [1,n]^k\}$ is a basis of $T^k(V^*)$ such that if $T \in T^k(V^*)$ then $T = \sum_{I \in [1,n]^k}T(\vec{e}_I)x_I$.

But now if $T \in \Lambda^k(V^*)$ then, as noted above, $T(\vec{e}_I)=0$ unless the vectors $e_{i_j}$ are pairwise distinct, or equivalently, $I \colon [1,k]\to [1,n]$ is injective. Now the symmetric group $S_k$ acts on $[1,n]^k$ by precomposing $\sigma(I) =I \circ \sigma^{-1}$, and it is easy see that $\sigma(x_I) = x_{\sigma(I)}$. Thus if $I\in [1,n]^k$ is injective, there is a unique $\sigma \in S_k$ such that $\sigma(I)$ is strictly increasing. It follows that if we let $\mathcal P(k,n) = \{S \subseteq [1,n]: |S|=k\}$ be the set of all $k$-element subsets of $[1,n]$ and note that we may embed $\mathcal P(k,n)$ into $[1,n]^k$ by identifying a $k$-element subset $S$ of $[1,n]$ with the $k$-tuple given by listing it elements in increasing order, then any $T \in \Lambda^k(V^*)\subseteq T^k_{\varepsilon}(V^*)$ can be written as $$ T = \sum_{S\in \mathcal P(k,n)} T(\vec{e}_S)\sum_{\sigma \in S_k} \varepsilon(\sigma)x_{\sigma(S)} $$ Thus if we set $a_k = \sum_{\sigma \in S_k} \varepsilon(\sigma)\sigma \in \mathsf k[S_k]$ the group algebra of $S_k$ and $\omega_S: = a_k(x_S)$ it follows that $\Lambda^k(V^*)$ lies in the span of $\{\omega_S: S\subseteq [1,n], |S|=k\}$, where $T\in \Lambda^k(V^*)$ satisfies $T = \sum_{S\in \mathcal P(k,n)}T(\vec{e}_S)\omega_S$.

On the other hand, we claim $\mathrm{im}(a_k)\subseteq \Lambda^k(V^*)$ and hence in fact $\mathrm{im}(a_k) = \Lambda^k(V^*)$ and in particular $\{\omega_S: S \subseteq [1,n],|S|=k\}$ is a basis of $\Lambda^k(V^*)$. (The set $\{\omega_S: S \in \mathcal P(k,n)\}$ is linearly independent since, for example, the set $\{x_S: S \in \mathcal P(k,n)\}$ is.) Thus $$ \dim(\Lambda^k(V^*) = \dim(\bigwedge\!{^k(}V) = \binom{n}{k}. $$

To see that $\mathrm{im}(a_k)\subseteq \Lambda^k(V^*)$, suppose that $\vec{v} = (v_1,\ldots,v_k)\in V^k$ has $v_i=v_{i+1}$ then for any $T\in T^k(V^*)$ one has $\sigma_i(T)(v_1,\ldots,v_k) = T(v_1,\ldots,v_k)$ so that $(e-\sigma_i)(T)(v_1,\ldots,v_k)=0$. But if $A_k = \{\sigma\in S_k: \varepsilon(\sigma)=1\}$ then $S_k = A_k \sqcup A_k\sigma_i$ and hence if $b_k= \sum_{\tau \in A_k} \tau$ we have $a_k = b_k(e-\sigma_i)$, hence $a_k(T)(v_1,\ldots, v_k) = b_k(e-\sigma_i)(T)(v_1,\ldots,v_k) =0$ since $(e-\sigma_i)(T)(v_1,\ldots,v_k) =0$ whenever $v_i=v_{i+1}$.

Duality

Now $\bigwedge(V)$, as a quotient of $T(V)$, is naturally an associative algebra, whereas the space of alternating forms $\Lambda(V^*)$ is a subspace of $T(V^*)$ which is never a subalgebra of $T(V^*)$ (assuming $\dim(V)>0$) (regardless of the characteristic of $\mathsf k$).

However, using ($\dagger$) it follows that $$ \langle b, a_k(\beta)\rangle = \langle b, \sum_{\sigma\in S_k} \epsilon(\sigma)\sigma(\beta)\rangle = \langle \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma^{-1}(b),\beta \rangle = \langle a_k(b),\beta\rangle, $$ where the final equality holds because $\epsilon(\sigma)= \epsilon(\sigma^{-1})$. In particular, since $J_k^0=\Lambda^k(V^*) = a_k(T^k(V^*))$ we see that $b \in J_k$ if and only if $\langle b, a_k(x_I)\rangle =0$ for all $I \in [1,n]^k$, hence $\langle a_k(b),x_I\rangle=0$ for all $I$, and hence $a_k(b)=0$. It follows that $J_k = \ker(a_k\colon T^k(V)\to T^k(V))$ for all $k$.

Now applying the above to $V^*$ and the ideal $\mathcal J$ of $T(V^*)$ generated by $\{f\otimes f: f \in V^*\}$ (identifying $V\cong V^{**}$ since $V$ is finite-dimensional), we see that if $a=\bigoplus_{k\geq 0} a_k$, a graded linear endomorphism of $T(V^*)$, then $N:=\ker(a)= \bigoplus_{k \geq 0} N_k$ where $N_k = \ker(a_k\colon T^k(V^*)\to T^k(V^*))$, then $N=\mathcal J$ is an ideal of $T(V^*)$. Since we have also seen that $\mathrm{im}(a) = \Lambda(V^*)$ the following trivial Lemma gives $\Lambda(V^*)$ an algebra structure:

Lemma: Let $A$ be a graded algebra and suppose that $p\colon A\to A$ is a graded linear map such that $J=\ker(p)$ is graded ideal of $A$. Let $C=p(A)$. Then there is a unique algebra structure on $C$ such that the induced map $\bar{p}\colon A/J \to C$ is an isomorphism of algebras. Indeed if the product on $C$ is denoted $\wedge$ then it is given by $p(x)\wedge p(y) = p(xy), \forall x,y \in A$.

Applying the Lemma to $\mathcal T(V^*)$ equipped with the graded map $a \colon \mathcal T(V^*) \to \mathcal T(V^*)$ which is equal to $a_k$ on $T^k(V^*)$ equips $\Lambda(V^*)$ with an algebra structure.

Shuffle product: It might appear that the characterization $p(x)\wedge p(y) = p(xy)$ has the disadvantage that, given $c_1,c_2\in C$, one must choose lifts $x_1$ and $x_2$ such that $p(x_1)=c_1,p(x_2)=c_2$ in order to compute $c_1\wedge c_2 = p(x_1\otimes x_2)$.

Now $a_k^2 = k!a_k$, and hence if $\text{char}(\mathsf k)>\dim(V)$ one can replace $a$ by $\tilde{a}_k$ where $\tilde{a}_k:= a_k/k!$, so that $\tilde{a}_k^2=\tilde{a}_k$ take $x_i=c_i=\tilde{a}(c_i)$.

This rescaling strategy fails when $\text{char}(\mathsf k)=p\leq k$, when $k!=0 \in \mathsf k$, but there is an alternative approach which works uniformly: If $c_1 \in \Lambda^k(V^*)$ and $c_2 \in \Lambda^l(V^*)$ then note that we may embed $S_k\times S_l$ in $S_{k+l}$ as $S_{k,l}:=\text{Sym}([1,k])\times \text{Sym}([k+1,k+l])$. Let $a_k = \sum_{\sigma_1\in \text{Sym}([1,k])} \varepsilon(\sigma_1)\sigma_1$, and $a^k_l=\sum_{\sigma_2 \in \text{Sym}([k+1,k+l])}\varepsilon(\sigma_2)\sigma_2$. Next let $$ T_{k,l} = \{\pi \in S_{k+l}: \pi(1)<\pi(2)<\ldots<\pi(k),\pi(k+1)<\ldots<\pi(k+l)\} $$ be the $(k,l)$-shuffles in $S_{k+l}$. Then $T_{k,l}$ is a tranversal for the cosets of $S_{k,l}$, that is, $S_{k+l} = \bigsqcup_{\tau \in T_{k,l}}\tau S_{k,l}$. It follows that if we set $t_{k,l} =\sum_{\tau \in T_{k,l}}\varepsilon(\tau)\tau$ then $a_{k+l} = t_{k,l} a_k a^k_l$. Thus if $c_1=a_k(x_1)$ and $c_2 =a_l(x_2)$ then $$ c_1\wedge c_2 = a_{k+l}(x_1\otimes x_2) = t_{k,l}a_ka^k_l(x_1\otimes x_2) = t_{k,l}(a_k(x_1)\otimes a_l(x_2)) = t_{k,l}(c_1\otimes c_2). $$

Thus we see that, for $V$ a finite-dimensional vector space over any field $\mathsf k$, $\Lambda(V^*)$ is an associative algebra with product given by $$ (c_k\wedge c_l)(v_1,\ldots,v_{k+l}) = \sum_{\pi \in T_{k,l}}\varepsilon(\pi)c_k(v_{\pi(1)},\ldots,v_{\pi(k)})c_l(v_{\pi(k+1)},\ldots,v_{\pi(k+l)}) $$ for all $c_k\in \Lambda^k(V^*), c_l \in \Lambda^l(V^*)$, and $a= \bigoplus_{k \geq 0} a_k$ induces an isomorphism of algebras $\bar{a}\colon \bigwedge(V^*) \to \Lambda(V^*)$.