Let $(V, \langle .,. \rangle)$ be an euclidean vector space. For a linear map $F: V \to V$ we define $|F|^2 = \sum_i |F(e_i)|^2$ for an orthonormal basis $\{e_i\}$. We write $\hat{F} : \Lambda^2 V \to \Lambda^2 V, \quad \hat{F}(X \wedge Y) = F(X) \wedge Y + X \wedge F(Y).$

Then we want t show that $|\hat{F}|^2 = 2|F|^2 - \operatorname{tr}(F)^2.$

We have \begin{align*} |\hat{F}|^2 &= \sum_{i<j} |\hat{F}(e_i \wedge e_j)|^2 = \frac12 \sum_{i,j} |\hat{F}(e_i \wedge e_j)|^2 \\ &= \frac12 \sum_{i,j} \langle F(e_i) \wedge e_j + e_i \wedge F(e_j), F(e_i) \wedge e_j + e_i \wedge F(e_j) \rangle \\ &= \frac12 \sum_{i,j} |F(e_i) \wedge e_j|^2 + |e_i \wedge F(e_j)|^2 + 2\langle F(e_i) \wedge e_j, e_i \wedge F(e_j) \rangle \end{align*} With the induced inner product on $\Lambda^2 V$ we get \begin{align*} &|F(e_i) \wedge e_j|^2 = |F(e_i)|^2 - \langle F(e_i), e_j \rangle^2\\ &|e_i \wedge F(e_j)|^2 = |F(e_j)|^2 - \langle F(e_j), e_i \rangle^2 \\ &2\langle F(e_i)\wedge e_j, e_i \wedge F(e_j) \rangle = 2\langle F(e_i), e_i \rangle \langle e_j, F(e_j) \rangle - 2\langle F(e_i), F(e_j) \rangle \delta_{ij} \end{align*} so \begin{align*} |\hat{F}|^2 =\frac12 \left( (\sum_j |F|^2) + (\sum_i |F|^2) + 2 (\operatorname{tr} F)^2 - 2|F|^2 - 2\sum_{i,j} \langle F(e_j), e_i \rangle^2 \right) \end{align*}

Is that correct? How do I proceed?

  • $\begingroup$ Your formula simplifies to $(n-2)|F|^2+(\text{tr}\, F)^2$, since $\sum\limits_{i,j} \langle F(e_j),e_i\rangle^2 = |F|^2$. This formula checks when $n=2$, whereas their formula definitely does not. For $n=2$, $\hat F = \text{tr}\,F$. $\endgroup$ – Ted Shifrin Jul 5 '18 at 21:19
  • $\begingroup$ @Ted Thanks, can you explain, why $\hat{F} = \operatorname{tr} F$ for $n=2$? $\endgroup$ – blat Jul 5 '18 at 21:45
  • $\begingroup$ Just write it out. If $F=[a_{ij}]$ with respect to the standard orthonormal basis, then $\hat F(e_1\wedge e_2) = F(e_1)\wedge e_2 + e_1\wedge F(e_2) = (a_{11}+a_{22})e_1\wedge e_2$. $\endgroup$ – Ted Shifrin Jul 5 '18 at 21:47

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