# Cauchy Schwarz inequality on $V \wedge V$

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?

• How is this an inner product? – Jacky Chong Oct 19 '16 at 16:12
• @JackyChong it is a norm, you can extract the inner product using the parallelogram: $$\langle a, b \rangle = \frac{1}{4}\big(|| a + b ||^2 - ||a - b||^2\big)$$ – cactus314 Oct 19 '16 at 17:17
• But $|v\wedge v|^2 = 0$. – Jacky Chong Oct 19 '16 at 20:09
• @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? $$\langle v_1 \wedge v_2, w_1 \wedge w_2 \rangle_2 \leq ||v_1 \wedge v_2||_1 \; ||w_1 \wedge w_2||_2$$ – cactus314 Oct 19 '16 at 20:19
• Okay. My mistake. I completely understand. So what's your question since you already understand $\langle v_1\wedge v_2, w_1\wedge w_2\rangle \leq \| v_1\wedge v_2\|_2 \|w_1\wedge w_2\|_2$? – Jacky Chong Oct 19 '16 at 20:33