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

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.

• How are you defining your coefficients? $$\omega = \omega_{i_1\cdots i_k}e^{i_1}\wedge\dotsb\wedge e^{i_k}$$ or $$\omega = \frac1{k!}\omega_{i_1\cdots i_k}e^{i_1}\wedge\dotsb\wedge e^{i_k}$$? Mar 15, 2023 at 22:35
• The former, since when I use einstein summation convention, I use the unstandard convention of only summing over ordered multi indices. The only time I don't sum over ordered multi indices is when I write a $\sum_{i_1\cdots i_k}$ with not less than signs. Mar 16, 2023 at 0:12
• I don't think that's a standard convention. But anyway, where you define $\langle\omega,\eta\rangle$, am I right that what you're saying is you took the canonical inner product $\langle\cdot,\cdot\rangle_T$ on $T^{0,k}$ $$\langle v^1\otimes\dotsb\otimes v^k, w^1\otimes\dotsb\otimes w^k\rangle = (v^1\cdot w^1)\dotsb(v^k\cdot w^k)$$ and defined $\langle\omega,\eta\rangle = \frac1{k!}\langle\omega,\eta\rangle_T$? Mar 16, 2023 at 1:36
• Precisely. And no I agree it is not standard convention at all, but I hate summation signs, and didnt want to double count. Thats why I called it unstandard convention Mar 16, 2023 at 2:00
• Oh, wow, you said unstandard. Apologies, I read that wrong! Mar 16, 2023 at 2:46

$$\newcommand\sgn{\mathrm{sgn}} \newcommand\alt{\mathrm{alt}} \newcommand\AltExt{\mathop\Lambda} \newcommand\Ext{\mathop{\textstyle\bigwedge}}$$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.

• Why does the lack of a normalization factor make the algebra non associative? I thought that it was just convention? Perhaps it’s a little messy, but I had the wedge product of two homogenous elements of order $k$ and $l$ have a factor of $1/k!l!$, and it all seemed to workout with the $v_1\wedge\cdots\wedge v_k$ defined as that sum above. I might’ve made a computational mistake though. Mar 10, 2023 at 2:10
• sorry for all the bother, and thank you for all your help today, but please see the edit for a more explicit explanation of my comment. Mar 10, 2023 at 3:57
• Hi again, I made a post detailing my main confusion here, as this post got unfocused very quickly. Mar 12, 2023 at 1:53

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.)

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 $$j_1 in $$[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)}. 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 $$i_1, we set $$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 $$S_1\cap S_2 = \emptyset$$.