Exterior algebra as a quotient is the same as exterior algebra as a vector subspace of the tensor algebra.

Question

I am writing an expository paper, and in it I defined $\Lambda^k(V)$ as the subspace of alternating tensors of order $k$, i.e. as a vector subspace of $V^{\otimes^k}$, where $V$ is a $\mathbb{K}=\mathbb{C},\mathbb{R}$-linear vector space. I then defined the exterior algebra with the wedge product as the graded algebra:

$$\Lambda(V)=\bigoplus_{k=0}^n\Lambda^k(V)$$

I am now developing the exterior algebra of vector space as a quotient of the tensor algebra: $$T(V)=\bigoplus_{n=0}^\infty V^{\otimes ^n}$$ with the ideal $I\subset T(V)$ generated by: $$\{v\otimes v|v\in V\}$$ My question what is sufficient to show that these are the same algebras? I had already proved that of a graded algebra by an ideal generated by homogenous elements is graded, and that $T(V)/I$ satisfies:

$$T(V)/I=\bigoplus T(V)_k/(I\cap T(V)_k)$$ And furthermore that:

$$T(V)_0/(I\cap T(V)_k=\mathbb{K}\qquad \text{and}\qquad T(V)_1/(I\cap T(V)_1)=V$$

I have also shown that for $k>n$ that $T(V)_k/(I\cap T(V)_k)=\{0\}$, and that multiplication in $T(V)/I$ satisfies: $$ \begin{align} [v\otimes v]=&0\\ [v\otimes w]=&-[w\otimes v] \end{align} $$ for all $v,w \in V$.

I am however really struggling to show that $T(V)_k/I_k=\Lambda^k(V)$. My thought process was to write the surjective linear map:

$$\phi:T(V)_k\longrightarrow \Lambda^k(V)$$

which on simple tensors is given by:

$$v_1\otimes\cdots\otimes v_k\longmapsto \sum_{\sigma\in S^k}\text{sgn}(\sigma)v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(k)}$$ and then show that $I\cap T(V)_k=\ker \phi$, so that I could apply a decomposition theorem I had proven earlier. However, showing that $\ker\phi\subset I\cap T(V)_k$ has proven near impossible. I know it should be true, but in the case where $a\in\ker\phi$ is not a simple tensor I can't seem to figure out the argument.

So my question is this, how do we see that these two constructions are equivalent? Is it necessary to go the route I am going, or am I making life harder for myself? Is it enough to just show that product in $T(V)/I$ satisfies the same properties of the wedge product in $\Lambda(V)$?

Edit: So basically this how I was doing things earlier, and it was in order to motivated differential forms, so everything was done with a dual basis. Let $V=\mathbb{R}^n$, $\{e_i\}$ be the standard basis, and $\{e^i\}$ be the dual. I defined a simple $k$ form as:

$$e^1\wedge \cdots \wedge e^k=\sum_{\sigma\in S^k}\text{sgn}(\sigma)e^{\sigma(1)}\otimes \cdots\otimes e^{\sigma(k)} $$

And then I said that the wedge product of $k$ form $\omega$, and $l$ form $\eta$ was given by it's action on vectors $v_1,\dots, v_{k+l}$: $$(\omega\wedge\eta)(v_1,\cdots,v_{k+l})=\frac{1}{k!l!}\sum_{\sigma\in S^{k+l}}\omega(v_{\sigma(1)},\dots, v_{\sigma(k)})\cdot \eta(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})$$

I liked these for two reasons, first off, if I let $V=\mathbb{R}^3$, $\omega=e^1\wedge e^2$, then with $v_1=a^ie_i$, $v_2=b^ie_i$, and $v_3=c^ie_i$: $$ \begin{align} (\omega\wedge e^3)(v_1,v_2,v_3)=&\frac{1}{2}\sum_{\sigma\in S^3}\text{sgn}(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})e^3(v_{\sigma(3)})\\ =&\frac{1}{2}\left(\omega(v_1,v_2)e^3(v_3)+\omega(v_2,v_3)e^3(v_1)+\omega(v_3,v_1)e^3(v_2)\right.\\ &-\left.\omega(v_2,v_1)e^3(v_3)-\omega(v_1,v_3)e^3(v_2)-\omega(v_3,v_2)e^3(v_1) \right)\\ =&\omega(v_1,v_2)e^3(v_3)+\omega(v_2,v_3)e^3(v_1)+\omega(v_3,v_1)e^3(v_2)\\ =&(a^1b^2-a^2b^1)c^3+(b^1c^2-b^2c^1)a^3+(c^1a^2-c^2a^1)b^3\\ =&\det(v_1,v_2,v_3) \end{align} $$ while: $$ \begin{align} e^1\wedge e^2\wedge e^3(v_1,v_2,v_3)=&\sum_{\sigma\in S^3}\text{\sgn}(\sigma)e^{\sigma(1)}\otimes e^{\sigma(2)}\otimes e^{\sigma(3)}(v_1,v_2,v_3)\\ =&a^1b^2c^3+b^1c^2a^3+c^1a^2b^3-b^1a^2c^3-a^1c^2b^3-c^1b^2a^3\\ =&\det(v_1,v_2,v_3) \end{align} $$ The fact these two line up felt important to me, since $\omega\wedge e^3=e^1\wedge e^2\wedge e^3$, and I am pretty sure if I add a $1/3!$ then these two things will be different, something I did not want. Secondly, when defining an inner product on $T^{0,k}$ tensors, I wanted the restriction to the subspace $\Lambda^k(V)$ to basically have a factor of $k!$ so I could rescale the inner product by $1/k!$ and obtain the standard formula: $$\langle \omega,\eta\rangle=\frac{1}{k!}\sum_{i_1\cdots i_k}\omega_{i_1 \cdots i_k}\eta^{i_1\cdots i_k}$$

Edit #2:

I see that what I am doing is primarily used in differential geometry, as my wedge product in terms of the Alt, carries a $(k+l)!/k!l!$ term. Does this factor make the two algebra's not isomorphic to one another? Is there a way around this? And why does: $$e^1\wedge \cdots\wedge e^k=\sum_{\sigma\in S^k}\text{sgn}(\sigma)e^{\sigma(1)}\otimes \cdots\otimes e^{\sigma(k)}$$ match what I am doing so well? In Lee's smooth manifolds, he has that $e^1\wedge \cdots \wedge e^k$ is the determinant of the $k\times k$ submatrix consisting of columns of the first $k$ components of $k$ vectors, and the definition above matches that perfectly, but I don’t quite see how to reconcile this with the wedge product as defined above.

I am using this construction as a way to motivate the clifford algebras as a deformation of the wedge product, so it would be quite sad for me if these two constructions are actually incompatible.

Edit 3:

I’m sorry it seems this post has gone all over the place, perhaps I will split it into separate questions, if people think that is wise.

However, I do believe I know how to reconcile what I wrote for for the simple $k$-form. Let $\{e^i\}$ be the dual basis for $V$, we want to show that: $$ e^{i_1}\wedge \cdots \wedge e^{i_k}=\sum_{\sigma\in S_k}\text{sgn}(\sigma)e^{\sigma(i_1)}\otimes \cdots \otimes e^{\sigma(i_k)} $$ We proceed by induction, the $1$st case is trivial, so we assume the $k$th case and apply the definition of the wedge product: $$ \begin{align} (e^{i_1}\wedge \cdots \wedge e^{i_k})\wedge e^{i_{k+1}}(v_1,\cdots ,v_{k+1})=&\frac{1}{k!}\sum_{\sigma\in S^{k+1}}\text{sgn}(\sigma)e^{i_1}\wedge \cdots \wedge e^{i_k}(v_{\sigma(1)},\cdots, v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)})\\ =&\frac{1}{k!}\sum_{\sigma\in S^{k+1}}\sum_{\tau\in S^k}\text{sgn} (\sigma)\text{sgn}(\tau) e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)}) \end{align} $$ For each $\sigma$ there are $k!$ factorial $\sigma'$'s satisfying $\sigma(k+1)=\sigma'(k+1)$, including $\sigma$. We can then split $S^{k+1}$ into $k+1$ sets, each consisting of the of the permutations which satisfy the aforementioned property. Denote each set by $A^l$, then our sum can be written as: $$ \begin{align} ``\text{ }''=&\frac{1}{k!}\sum_{l=1}^{k+1}\sum_{\sigma \in A^l}\sum_{\tau \in S^k}\text{sgn} (\sigma)\text{sgn}(\tau) e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)}) \end{align} $$ Fix an $l$, such that $\sigma(k+1)=j$ then: $$ \begin{align} \sum_{\sigma \in A^l}&\sum_{\tau \in S^k}\text{sgn} (\sigma)\text{sgn}(\tau) e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)})\\ =&e^{i_{k+1}}(v_j) \sum_{\sigma \in A_i}\sum_{\tau \in S^k}\text{sgn} (\sigma)\text{sgn}(\tau) e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})\\ \end{align} $$ It doesn't matter computationally whether we permute the covectors, or the vectors, as summing over either gives us every combination of $e^{i_j}(v_l)$, where $1\leq j,l\leq k$. Hence, fixing $\tau$ and $\tau'$ in $S_{k}$ we claim that: $$ \begin{align} \sum_{\sigma \in A_l}\text{sgn} (\sigma)&\text{sgn}(\tau) e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})\\ =&\sum_{\sigma\in A_l}\text{sgn}(\sigma') \text{sgn}(\tau') e^{\tau(i_1)}(v_{\sigma'(1)})\cdots e^{\tau(i_k)}(v_{\sigma'(k)}) \end{align} $$ In the case where $\text{sgn}(\tau)=\text{sgn}(\tau')$ we have that $\tau$ and $\tau'$ differ by an even amount of swaps, so fore very $\sigma$, the unique $\sigma'$ which satisfies: $$e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})=e^{\tau'(i_1)}(v_{\sigma'(1)})\cdots e^{\tau'(i_k)}(v_{\sigma'(k)}$$ must also differ by an even amount swaps, implying that each term in left sum, is equal to some term in the right sum, implying the claim. A similar argument follows in the case where $\text{sng}(\tau)=-\text{sgn}(\tau')$. Since for each $\tau$ the above equality holds, we have that: $$ \begin{align} \sum_{\sigma \in A^l}&\sum_{\tau \in S^k}\text{sgn} (\sigma)\text{sgn}(\tau) e^{\tau(i_1)}(v_{\sigma(1)})\cdots e^{\tau(i_k)}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)})\\ =&k!\sum_{\sigma \in A^l}\text{sgn} (\sigma)e^{i_1}(v_{\sigma(1)})\cdots e^{i_k}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)}) \end{align} $$ implying that: $$ \begin{align} (e^{i_1}\wedge \cdots \wedge e^{i_k})\wedge e^{i_{k+1}}(v_1,\cdots ,v_{k+1})=&\sum_{l=1}^{k+1}\sum_{\sigma \in A^l}\text{sgn} (\sigma) e^{i_1}(v_{\sigma(1)})\cdots e^{i_k}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)})\\ =&\sum_{\sigma\in S^{k+1}}\text{sgn} (\sigma) e^{i_1}(v_{\sigma(1)})\cdots e^{i_k}(v_{\sigma(k)})\cdot e^{i_{k+1}}(v_{\sigma(k+1)})\\ =&\sum_{\sigma\in S^{k+1}}\text{sgn} (\sigma) e^{\sigma(i_1)}\otimes \cdots \otimes e^{\sigma(i_k)}\otimes e^{\sigma(i_{k+1})}(v_1,\cdots,v_{k+1}) \end{align} $$ implying the original claim as $(v_1,\cdots, v_{k+1})$ was an arbitrary set of vectors.

Answer 1

$ \newcommand\sgn{\mathrm{sgn}} \newcommand\alt{\mathrm{alt}} \newcommand\AltExt{\mathop\Lambda} \newcommand\Ext{\mathop{\textstyle\bigwedge}} $Your $\phi$ as defined will not work; the wedge product you get from this is non-associative. You have to introduce the normalization factor: $$ v_1\otimes\dotsb\otimes v_k = \color{red}{\frac1{k!}}\sum_{\sigma\in S^k}\sgn(\sigma)v_{\sigma(1)}\otimes\dotsb\otimes v_{\sigma(k)}. $$ (And its for this reason the the exterior algebra is not isomorphic to a subalgebra of $T(V)$ when working over a fields of nonzero characteristic.)

Let us denote $\Ext V = T(V)/I$ and identify $V$ with its image in $\Ext V$ under the canonical projection. What I think is going to be easiest is to realize that this definition of $\Ext V$ gives us a universal property:

• Let $A$ be any associative algebra. Then every linear $f : V \to A$ such that $f(v)^2 = 0$ for all $v \in V$ extends uniquely to an algebra homomorphism $f' : \Ext V \to A$ such that $f'(v) = f(v)$ for all $v \in V$.

Then we define a wedge product on $\AltExt(V)$ by defining the alternation map $\alt : T(V) \to \AltExt(V)$ grade-wise $$ \alt(v_1\otimes\dotsb\otimes v_k) = \frac1{k!}\sum_{\sigma\in S^k}\sgn(\sigma)v_{\sigma(1)}\otimes\dotsb\otimes v_{\sigma(k)} $$ whence $\AltExt(V) = \alt(T(V))$ and we define $$ X\wedge Y = \alt(X\otimes Y) $$ for any $X, Y \in \AltExt(V)$. By linearity it suffices to show this is associative on simple tensors. Once thats done $\AltExt(V)$ together with $\wedge$ is an associative algebra, and it is easy to see that the map taking $v \in V$ to itself in $\AltExt(V)$ satifies the premises of the universal property of $\Ext V$; hence there is an algebra homomorphism $\phi : \Ext V \to \AltExt(V)$ preserving vectors. Showing one of surjectivity or injectivity should not be difficult; if you do at least one, then it suffices to argue that $\Ext V$ and $\AltExt(V)$ have the same dimension, and we're done.

Answer 2

The two approaches you describe to the exterior algebra are probably best thought of as being dual to each other:

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 $I$ generated by $R$. If we set $I_k = T^k(V)\cap I$, then $$ \bigwedge(V) = \bigoplus_{k \geq 0}\bigwedge^k(V), \quad \text{where } \bigwedge^k(V) = T^k(V)/I_k. $$

On the other hand, if $T(V^*)$ is the tensor algebra of $V^*$, then we may view $t\in T^k(V^*)$ as a $k$-linear map $t\colon V^k \to \mathbb R$. We then define $\Lambda(V^*) = \bigoplus_{k\geq 0} \Lambda^k(V^*)$, the space of alternating multilinear forms on $V$, where, writing $[1,k] = \{1,2,\ldots,k\}$ for convenience of notation, $$ \Lambda^k(V^*) = \{t \in T^k(V^*): t(v_1,\ldots,v_k)=0 \text{ if } \exists i \in [1,k], v_i=v_{i+1}\}. $$

1. Alternating forms

We can analyse $\Lambda^k(V^*)$ using the symmetric group: Note that $S_k$, the symmetric group on $k$ letter, acts on $T^k(V^*)$, where if $\sigma \in S_k$ and $t \in T^k(V^*)$ then $$ \sigma(t)(v_1,\ldots,v_k) = t(v_{\sigma(1)},\ldots,t_{\sigma(k)}), \quad \forall v_1,\ldots,v_k \in V. $$ It is not immediately clear that the action of $S_k$ preserves the subspace $\Lambda^k(V^*)$. However, if $t\in \Lambda^k(V^*)$ and we have vectors $v_1,\ldots,v_k \in V$, writing $t_{i,i+1}(w_1,w_2)$ for $t(v_1,\ldots,v_{i-1},w_1,w_2,v_{i+2},\ldots,v_k)$ we have $$ \begin{split} 0&=t_{i,i+1}(v_i+v_{i+1},v_i+v_{i+1})\\ &= t_{i,i+1}(v_i,v_i) + t_{i,i+1}(v_i,v_{i+1})+t_{i,i+1}(v_{i+1},v_i)+ t_{i,i+1}(v_{i+1},v_{i+1})\\ &= 0+ t_{i,i+1}(v_i,v_{i+1}) + t_{i,i+1}(v_{i+1},v_i)+0. \end{split} $$ so that $t_{i,i+1}(v_i,v_{i+1}) = -t_{i,i+1}(v_{i+1},v_i)$. 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 $S_k$. Moreover, $S_k$ acts on $\Lambda^k(V^*)$ via the sign character $\epsilon$, where $\epsilon\colon S_k \to \{\pm 1\}$ is the unique homomorphism such that $\epsilon(\sigma_i) = -1$ for all $i \in \{1,\ldots, k-1\}$. (That such a homomorphism exists can be shown, for example, by considering the polynomial $\Delta = \prod_{1\leq i<j\leq k}(x_i-x_j)$.)

But now if $t \in T^k(V^*)$ is such that $\sigma(t) = \epsilon(\sigma).t$, then taking $\sigma=\sigma_i$ and writing $t_{i,i+1}$ as before we see that $$ -t_{i,i+1}(v_i,v_i) = \sigma_i(t_{i,i+1})(v_i,v_i) = t_{i,i+1}(v_i,v_i) $$ hence $t_{i,i+1}(v_i,v_i)=0$ (since $\text{char}(\mathbb R)\neq 2$) so that, since this holds for all $i\in [1,n-1]$, $t\in \Lambda^k(V^*)$. It follows that $\Lambda^k(V^*) = T^k_{\epsilon}(V^*)$ where we set $$ T^k_{\epsilon}(V^*):= \{t \in T^k(V^*): \sigma(t) = \epsilon(\sigma).t\}. $$

One advantage to the description of $\Lambda^k(V^*)$ as $T^k_{\epsilon}(V^*)$ is that it is easy to produce a basis of $\Lambda^k(V^*)$ using it: If $\{e_1,\ldots,e_n\}$ is a basis of $V$ and $\{x_1,\ldots,x_n\}$ the corresponding dual basis of $V^*$, then $T^k(V^*)$ inherits an induced basis $\{x_I: I \in [1,n]^k\}$, where if $I=(i_1,i_2,\ldots, i_k)$, with, for $1\leq j \leq k$, each $i_j \in [1,n]$, then we set $x_I = x_{i_1}\otimes\ldots\otimes x_{i_k}$, the $k$-linear map given by $$ x_{i_1}\otimes \ldots \otimes x_{i_k}(v_1,\ldots,v_k) = \prod_{j=1}^k x_{i_j}(v_j), \quad \forall v_1,\ldots,v_k \in V. $$

Thus given $t \in \Lambda^k(V^*)$ we may write $t = \sum_{I} a_Ix_I$. But now if $I$ is such that $i_{j_1}=i_{j_2}$ for some $j_1<j_2$ in $[1,k]$ then if $\sigma_{j_1,j_2}$ denotes the transposition in $S_k$ which interchanges $j_1$ and $j_2$, we have $\sigma_{j_1,j_2}(x_I)= x_I$. Since $\sigma_{j_1,j_2}(t)=-t$, it follows that if $a_I\neq 0$ then the elements of the tuple $i_{j_1},\ldots,i_{j_k}$ must all be distinct. In this case, however, there is a unique permutation $\sigma_I$ such that $i_{\sigma_I(1)}<i_{\sigma_I(2)}<\ldots<i_{\sigma_I(k)}$. Thus if we let $\mathcal S(n,k)$ denote the set of $k$-element subsets of $[1,n]$ and, for each $S \in \mathcal S(n,k)$ with $S=\{i_1,\ldots,i_k\}$ where $i_1<i_2<\ldots<i_k$, we set $x_S = x_{I_S}$ where $I_S =(i_1,i_2,\ldots,i_k)$ then it follows that $$ t= \sum_{S\in \mathcal S(n,k)} a_{I_S}\omega_S, \quad \text{ where} \quad \omega_S =\sum_{\sigma\in S_k} \epsilon_k(\sigma).\sigma(x_S) $$ and hence $\{\omega_S: S \in \mathcal S(n,k)\}$ is a basis of $\Lambda^k(V^*)$.

Now note that if we set $a_k = \sum_{\sigma\in S_k} \epsilon(\sigma).\sigma$ (an element of the group algebra of $S_k$) so that $\omega_S = a_k(x_S)$, the discussion above shows that for any $I \in [1,n]^k$, either there is some $\tau \in S_k$ and $S\in \mathcal S(n,k)$ such that $\tau(I)= I_S$, or there is a transposition $\sigma_{j_1,j_2}$ with $I = \sigma_{j_1,j_2}(I)$. We claim that $a_k(x_I) = \epsilon(\tau) \omega_S$ in the first case and $a_k(x_I)=0$ in the second. Indeed for any $\tau \in S_k$ we have $$ a_k \tau = \sum_{\sigma \in S_k} \epsilon(\sigma)\sigma\tau= \epsilon(\tau)\sum_{\sigma \in S_k}\epsilon(\sigma)\epsilon(\tau)\sigma\tau = \epsilon(\tau)\sum_{\rho \in S_k}\epsilon(\rho)\rho = \epsilon(\tau)a_k $$ where in the penultimate equality we set $\rho = \tau\sigma$. Thus if $\tau(I)=I_S$ so that $\tau(x_I)=x_{I_S}$, then $$ \omega_S = a_k(\tau(x_I))=(a_k\tau)(x_I) = \epsilon(\tau)a_k(x_I), $$ hence $a_k(x_I) = \epsilon(\tau)\omega_S$. On the other hand, if $\sigma_{j_1,j_2}(I)=I$ for some transposition $\sigma_{j_1,j_2}$ then we have $$ a_k(x_I) = a_k\sigma_{j_1,j_2}(x_I)= \epsilon(\sigma_{j_1,j_2})a_k(x_I)=-a_k(x_I) $$ so that $2a_k(x_I)=0$ and hence $a_k(x_I)=0$. It follows that a basis for $\ker(a_k)$ is given by $$\{x_I: a_k(x_I)=0\}\cup \{x_S -\epsilon_k(\sigma).\sigma(x_S): S\in \mathcal S(n,k), \sigma \in S_k\backslash\{e\}\}. $$ In particular, if we set $N_k = \ker(a_k)$, then using this basis and the basis we already obtained for $a_k(T^k(V^*))= \Lambda^k(V^*)$ we see that $N_k\cap \Lambda^k(V^*)=\{0\}$.

2. 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 $\dim(V)>0$). (Indeed the space of alternating forms is not a subalgebra of $T(V^*)$ in any characteristic.) However, under the natural pairing of $T^k(V^*)$ with $T^k(V)$, the subspace $\Lambda^k(V^*)$ is by definition precisely the annihilator of $I_k$, and hence it is naturally identified with $(\bigwedge^k(V))^*$.

Moreover, the actions of $S_k$ on $T^k(V)$ and $T^k(V^*)$ are compatible with the natural pairing between these spaces in the sense that $$ \langle \sigma(a),\sigma(\alpha)\rangle = \langle a,\alpha \rangle, \quad \forall a \in T^k(V),\alpha \in T^k(V^*), \sigma \in S_k. $$ Thus 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 $\Lambda^k(V^*) = a_k(T^k(V^*))$ we see that $b \in I_k$ 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 $I_k = \ker(a_k\colon T^k(V)\to T^k(V))$ for all $k$.

Now applying the above to $V^*$ and the ideal $J$ 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 $\ker(a)=N= \bigoplus_{k \geq 0} N_k$ where $N_k = \ker(a_k\colon T^k(V^*)\to T^k(V^*))$, then $N=J$ is an ideal of $T(V^*)$. Since we have also seen that $\mathrm{im}(a) = \Lambda(V^*)$ and $\ker(a)\cap \mathrm{im}(a)=\{0\}$ the following obvious Lemma gives $\Lambda(V^*)$ an algebra structure:

Lemma Suppose that $A$ is a graded algebra and $p\colon A\to A$ is a graded linear map satisfying

1. $I=\ker(p)$ is an ideal of $A$,
2. $p(A) \cap I =\{0\}$.

Let $C=p(A)$ so that $p$ induces a linear isomorphism $\bar{p}\colon A/I \to C$. Then there is a unique algebra structure $\wedge\colon C\times C \to C$ on $C$ such that $\bar{p}$ is an isomorphism of algebras, where $\wedge$ is given by $p(x)\wedge p(y) = p(xy), \forall x,y \in A$.

Thus if, for example, we set $\omega_S=a_k(x_S)$, and take $S_1 \in \mathcal S(n,k), S_2 \in \mathcal S(n,l)$ then $\omega_{S_1}\wedge \omega_{S_2} = a_{k+l}(x_{I_{S_1}}\otimes x_{I_{S_2}})$ which is $0$ if $S_1\cap S_2 \neq \emptyset$ and is $\pm \omega_{S_1\cup S_2}$ if $S_1\cap S_2 = \emptyset$.