Let $\mathcal{H}$ be a Hilbert space with orthonormal basis $\{\varphi_{k}\}_{k\in I}$. Take $\mathcal{H}^{\otimes n} := \overbrace{\mathcal{H}\otimes\cdots\otimes \mathcal{H}}^{\mbox{$n$ times}}$. An element of $\mathcal{H}^{\otimes n}$ can be expressed as: $$\psi = \sum_{\{k_{1},...,k_{n}\}\subset I}\alpha_{k_{1},...,k_{n}}(\varphi_{k_{1}}\otimes \cdots \otimes \varphi_{k_{n}})$$ with $\alpha_{k_{1},...,k_{n}} = \langle \varphi_{k_{1}}\otimes\cdots\otimes\varphi_{k_{n}},\psi\rangle$. Let us define $\sigma^{*}$ as an operator on $\mathcal{H}^{\otimes n}$ which acts on the basis elements as: $$\sigma^{*}(\varphi_{k_{1}}\otimes \cdots \otimes \varphi_{k_{n}}) := \varphi_{k_{\sigma(1)}}\otimes\cdots \otimes \varphi_{k_{\sigma(n)}}$$ where $\sigma$ is a permutation of the set $\{1,...,n\}$. We extend $\sigma^{*}$ to all $\mathcal{H}^{\otimes n}$ by linearity. Now, one can define: $$A_{n}:= \frac{1}{n!}\sum_{\sigma}\epsilon_{\sigma}\sigma^{*}$$ an antisymmetrization operator on $\mathcal{H}^{\otimes n}$. Here $\epsilon_{\sigma}$ is the sign of the associate permutation $\sigma$. Then $A_{n}$ is an orthogonal projection and, if $A_{n}\mathcal{H}^{\otimes n}$ denotes its range, the fermionic Fock space is defined to be: $$\mathcal{F}_{f}(\mathcal{H}) := \bigoplus_{n=0}^{\infty}A_{n}\mathcal{H}^{\otimes n}$$ with $A_{0}\mathcal{H}^{0} := \mathbb{C}$.

Alternatively, let us say that a tensor $\psi \in \mathcal{H}^{\otimes n}$ is antisymmetric if $\sigma^{*}\psi = \epsilon_{\sigma}\psi$ for every permutation $\sigma$. Take $\wedge^{n}\mathcal{H}$ to be the subspace of all antisymmetric tensors of $\mathcal{H}^{\otimes n}$ and $\wedge^{0}\mathcal{H} := \mathbb{C}$.

**Question:** Can I use the second approach to define fermionic Fock spaces in an equivalent way, as before? In other words, if I set $\mathcal{F}'_{f}(\mathcal{H}) := \bigoplus_{n=0}^{\infty}\wedge^{n}\mathcal{H}$, does it follow that $\mathcal{F}_{f}(\mathcal{H}) = \mathcal{F}'_{f}(\mathcal{H})$? Equivalently: is it possible to prove that every $\psi \in \wedge^{n}\mathcal{H}$ can be expressed as $\psi = \frac{1}{n!}\sum_{\sigma}\epsilon_{\sigma}\sigma^{*}\varphi$ for some $\varphi \in \mathcal{H}^{\otimes n}$?