the Cauchy Schwarz inequality says for example $$ (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) \geq (a_1 b_1+ a_2 b_2+a_3 b_3)^2$$ and with clever choices of $a$ and $b$ we can solve many problems. It can be read as saying $\cos \theta \leq 1$. It can be interpreted as quantifying the pigeonhole principle. Also it's related to the distance from a point to a line (leading to a proof). Half of real analysis is clever usages of this inequality.

Cauchy Schwartz works because it's related to Pythagoras theorem, and it extends to $n$ variables and to metric spaces. So here is my question. Let $V=\mathbb{R}^n$ with the dot product. Then $V \wedge V$ is also an inner product space with an induced inner product. How do I write the Cauchy Schwartz inequality there.

for Pythagoras theorem I have $$ |v \wedge w |^2= \sum (v_i w_j - w_i v_j)^2$$ which says the area squares of a parallelogram in $n$ space is the sum of the area squares of each projection into pairs of coordinate axes.

what is the analogue of Cauchy Schwartz?

  • $\begingroup$ How is this an inner product? $\endgroup$ – Jacky Chong Oct 19 '16 at 16:12
  • $\begingroup$ @JackyChong it is a norm, you can extract the inner product using the parallelogram: $$ \langle a, b \rangle = \frac{1}{4}\big(|| a + b ||^2 - ||a - b||^2\big) $$ $\endgroup$ – cactus314 Oct 19 '16 at 17:17
  • $\begingroup$ But $|v\wedge v|^2 = 0$. $\endgroup$ – Jacky Chong Oct 19 '16 at 20:09
  • $\begingroup$ @JackyChong there are two inner products. There is $(V , \langle \cdot, \cdot \rangle_1)$ with basis $\{ e_1, \dots, e_n\}$ and there is an inner product on the wedge space $ (V \wedge V, \langle \cdot, \cdot \rangle_2)$ with $e_i \wedge e_j , e_k\wedge e_l$ perpendicular unless $i = k$ and $j = l$ or unless $i = l$ and $j = k$. Cauchy Schwartz inequality could refer to: $$ \langle v, w \rangle_1 \leq ||v||_1 \; ||w||_1 $$ or the same with respect the second norm. Can you picture this? $$ \langle v_1 \wedge v_2, w_1 \wedge w_2 \rangle_2 \leq ||v_1 \wedge v_2||_1 \; ||w_1 \wedge w_2||_2 $$ $\endgroup$ – cactus314 Oct 19 '16 at 20:19
  • $\begingroup$ Okay. My mistake. I completely understand. So what's your question since you already understand $\langle v_1\wedge v_2, w_1\wedge w_2\rangle \leq \| v_1\wedge v_2\|_2 \|w_1\wedge w_2\|_2$? $\endgroup$ – Jacky Chong Oct 19 '16 at 20:33

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