2

Wikipedia's Fock Space entry says that the Fock space is the direct sum of tensor products of $H$. However, it is not represented as $\bigoplus_{n=0}^{\infty} H^{\otimes n}$, but rather as $\bigoplus_{n=0}^{\infty} S_\nu H^{\otimes n}$, where $S_\nu$ symmetrizes ($\nu=+1$) or anti-symmetrizes ($\nu = -1$) a tensor.

Now my question is, what is the difference between $\bigoplus_{n=0}^{\infty} S_+ H^{\otimes n}$ and $\bigoplus_{n=0}^{\infty} S^n(H)$, where the latter is a symmetric algebra? Similarly, what is the difference between $\bigoplus_{n=0}^{\infty} S_- H^{\otimes n}$ and $\bigoplus_{n=0}^{\infty} \bigwedge^n(H)$, where the latter is an exterior algebra?

Godfly666
  • 33

1 Answers1

2

There are two ways of looking at the symmetrization of $H$:

  • as the quotient $S^n(H) = H^{\otimes n}/I$, where $I$ is the ideal generated $I$ by sums of the form $v\otimes w - w\otimes v$;
  • as the image of the operator $S_+:H^{\otimes n}\to H^{\otimes n}$ defined by sending generators to their symmetric product $$ S_+(v_1\otimes\cdots\otimes v_n) = \frac{1}{n!}\sum_{} v_{\sigma_1}\otimes\cdots\otimes v_{\sigma_n}, $$ where the summation runs over the permutations $(1,\dots,n)$.

The isomorphism $S^n(H) \cong S_+(H)$ follows from the 1st isomorphism theorem by showing that $\ker S_+$ is $I$.

Similarly, the exterior algebra can be defined either as the quotient $\bigwedge^n(H) = H^{\otimes n}/I$ by the ideal generated by sums of the form $v\otimes w+ w\otimes v$, or as the image of the operator $S_-:H^{\otimes n}\to H^{\otimes n}$ defined by sending generators to their antisymmetric product $$ S_-(v_1\otimes\cdots\otimes v_n) = \frac{1}{n!}\sum_{}\text{sign}(\sigma) v_{\sigma_1}\otimes\cdots\otimes v_{\sigma_n} =: v_1\wedge \cdots\wedge v_n, $$ and likewise we can obtain the isomorphism $\bigwedge^n(H)\cong S_-(H)$

Daniel Teixeira
  • 2,165