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This is my second attempt, following the first question here: Hilbert space with a "norm" that does not respect the triangle inequality?

I believe this inner product produces a Hilbert space, except its norm does not respect the triangle inequality. All axioms of the Hilbert space are respected. I prove this explicitly here. Is the space still a Hilbert space?

$u$ and $v$ are vectors of $2\times 2$ matrices $G$ from the orientation-preserving general linear group $GL^+(2,\mathbb{R})$ : i.e. $\det G > 0$.

$$ \langle u,v\rangle=\frac{1}{4}(||u+v||^2-||u-v||^2) $$

where

$$ ||x||^2=\sum_{G \in x}\det G $$

Where the notation $G\in x$ designate picking each element of the vector $x$ (which are $n\times n$ matrices).

Here, for the proofs, I am assume vectors have only one elements, but it generalizes to multiple elements. According to https://mathworld.wolfram.com/InnerProduct.html, the axioms of an inner product are as follows.

  1. $\langle u+v, w \rangle = \langle u,w\rangle + \langle v, w\rangle$

$$ \begin{align} &>u=\left( \begin{array}{cc} \text{u1} & \text{u2} \\ \text{u3} & \text{u4} \\ \end{array} \right) \\ &>v=\left( \begin{array}{cc} \text{v1} & \text{v2} \\ \text{v3} & \text{v4} \\ \end{array} \right) \\ &> w=\left( \begin{array}{cc} \text{w1} & \text{w2} \\ \text{w3} & \text{w4} \\ \end{array} \right)\\ &>\text{InnerProduct}(\text{a$\_$},\text{b$\_$})=\frac{1}{4} (| a+b| -| a-b| )\\ &>\text{Simplify}[\text{InnerProduct}(u+v,w)=\text{InnerProduct}(u,w)+\text{InnerProduct}(v,w)]\\ &out = true \end{align} $$

My apologies for using Mathematica for a "proof", but the notation would be substantial otherwise.

  1. $\langle \alpha v, w \rangle = \alpha \langle v,w \rangle$

$$ \begin{align} &>\text{Simplify}[\text{InnerProduct}(a u,v)=a \text{InnerProduct}(u,v)]\\ &out =true \end{align} $$

  1. $\langle v, w \rangle = \langle w,v \rangle$

$$ \begin{align} &>\text{Simplify}[\text{InnerProduct}(u,w)=\text{InnerProduct}(w,u)]\\ &out=true \end{align} $$

  1. $\langle u,u \rangle \geq 0 \text{ iff } u=0$

Here we have a bit of a pickle, but it's resolvable. Since, I have stated we only have $\det G > 0$, we can either state that the space has no $0$ in it which is still compliant with the axiom (it doesn't require a zero, it just takes that if there is one, then its inner product must be $0$). Or we can just manually add the zero-valued matrix to complete the space with a $0$.

$$ \langle u,u\rangle=||u||^2=\sum_{G \in u} \det G $$

Since the matrices $G$ have $\det g>0$, then it follows that $\langle u,u\rangle>0$.

Now, my concern is that the norm induced by the inner product does not seem to respect the triangle inequality (but maybe I am wrong and it does?).

$$ ||u||=\sqrt{\langle u,u \rangle} $$

And the triangle inequality is:

$$ ||u+v||\leq ||u||+||v|| $$

Yielding:

$$ \begin{align} &>\text{ExpandAll}[\text{InnerProduct}(u+v,u+v)]\\ &out=\text{u1} \text{u4}+\text{u1} \text{v4}-\text{u2} \text{u3}-\text{u2} \text{v3}-\text{u3} \text{v2}+\text{u4} \text{v1}+\text{v1} \text{v4}-\text{v2} \text{v3}\\ &>\text{ExpandAll}[\text{InnerProduct}(u,u)+\text{InnerProduct}(v,v)]\\ &out=\text{u1} \text{u4}-\text{u2} \text{u3}+\text{v1} \text{v4}-\text{v2} \text{v3} \end{align} $$

Can we show that:

$$ \begin{align} \text{u1} \text{u4}+\text{u1} \text{v4}-\text{u2} \text{u3}-\text{u2} \text{v3}-\text{u3} \text{v2}+\text{u4} \text{v1}+\text{v1} \text{v4}-\text{v2} \text{v3}&\leq \text{u1} \text{u4}-\text{u2} \text{u3}+\text{v1} \text{v4}-\text{v2} \text{v3}\\ \text{u1} \text{v4}-\text{u2} \text{v3}-\text{u3} \text{v2}+\text{u4}\text{v1} &\leq0 \end{align} $$

It appears to me that because the inequality contains a mixture of positive and negative signs, then it must that that the inequality fails. However, since it adheres to the axioms of the inner product, it must be that this is a norm. Are the restriction to the orientation-preserving general linear group "conspire" in some subtle manner to guarantee this inequality holds?

Anon21
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  • Can you explain your formula a bit more? You originally state that $u$ and $v$ are in $GL^+(2,\mathbb{R})$, but then how is the inner product defined for other matrices? – Michael Burr Jun 06 '21 at 12:49
  • @MichaelBurr For other matrices, the inner product would not be positive-definite (axiom 4), therefore I just don't consider them part of the domain at all. Let me know if this answers your question. – Anon21 Jun 06 '21 at 12:50
  • But, $GL^+(2,\mathbb{R})$ isn't closed under addition/subtraction (and it doesn't have $0$). So, what Hilbert space are you working on? – Michael Burr Jun 06 '21 at 12:54
  • @MichaelBurr You are suggesting that $GL^+(2,\mathbb{R})$ is not a vector space? Looking at the 8 axioms here: https://mathworld.wolfram.com/VectorSpace.html, Axioms 3 and 4 talk about a zero? So it seems I need to complete $GL^+(2,\mathbb{R})$ with the zero matrix, but I do not see how those 8 axioms implies completeness over subtraction. – Anon21 Jun 06 '21 at 13:01
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    Consider the matrices $\begin{pmatrix}2&0\0&2\end{pmatrix}$ and $\begin{pmatrix}-3&0\0&-1\end{pmatrix}$. If I understand your setup correctly, these are both in your set. But their sum is $\begin{pmatrix}-1&0\0&1\end{pmatrix}$, which is not. This is not a vector space since it is not closed under addition. The first sentence of the mathworld site has two additional properties: "closed under finite vector addition and scalar multiplication". – Michael Burr Jun 06 '21 at 13:04
  • @MichaelBurr What about this; say I open up the space to all $2\times 2$ matrices, then it is a vector space, but instead of an inner product, I get a degenerate bilinear form? – Anon21 Jun 06 '21 at 13:58
  • It might make sense to ask a new question describing what you’re looking for. As for the new example, note that if $u$ has negative determinant, then $\langle u,u\rangle$ is negative. – Michael Burr Jun 06 '21 at 14:44

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