# Representation of an arbitrary element on a fermionic Fock Space

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 $$\varphi \in \mathcal{H}^{\otimes n}$$?

If $$\psi\in\wedge^n\mathcal{H}$$ then by definition $$\sigma^*\psi = \epsilon_\sigma \psi$$ for each permutation $$\sigma$$ and so as $$\epsilon_\sigma \in \{\pm 1\}$$ we have $$A_n\psi = \frac{1}{n!} \sum_\sigma \epsilon_\sigma \sigma^*\psi = \frac{1}{n!} \sum_\sigma \epsilon_\sigma \epsilon_\sigma \psi = \psi.$$
Actually, the converse is much more interesting, as it really involves a little bit of representation theory. By the converse, I mean: show that if a tensor is in the range of $$A_n$$ then it is anti-symmetric. So, $$\psi = A_n\varphi$$ for some arbitrary $$\varphi$$. Then for a permutation $$\sigma$$, $$\sigma^*\psi = \frac{1}{n!} \sum_\tau \epsilon_\tau \sigma^*\tau^*\varphi.$$ Set $$\rho = \tau\sigma$$ and notice that $$\sigma^*\tau^*(\otimes_i \varphi_{k_i}) = \sigma^*(\otimes_i \varphi_{k_{\tau(i)}}) = \sigma^*(\otimes_i \varphi_{l_i})$$ say, where thus $$l_i = k_{\tau(i)}$$. Then $$l_{\sigma(i)} = k_{\tau(\sigma(i))}$$ and so $$\sigma^*\tau^*(\otimes_i \varphi_{k_i}) = \otimes_i \varphi_{k_{\tau\sigma(i)}} = (\tau\sigma)^*(\otimes_i \varphi_{k_i})$$. Thus we have an anti-representation of the symmetric group. As $$\epsilon$$ is a group homomorphism, $$\epsilon_\tau = \epsilon_{\rho\sigma^{-1}} = \epsilon_\rho\epsilon_\sigma$$. Thus $$\sigma^*\psi = \frac{1}{n!} \sum_\rho \epsilon_\rho\epsilon_\sigma \rho^*\varphi = \epsilon_\sigma A_n\varphi = \epsilon_\sigma \psi.$$ So $$\psi\in\wedge^n\mathcal{H}$$.