# The “ultimate” indefinite inner product space

This can be considered as a relative of Splitting a space into positive and negative parts.

Is there a real (non-trivial) vector space $V$, endowed with a nondegenerate symmetric bilinear pairing $\langle-,-\rangle : V^2 \to \mathbb{R}$, satisfying the following property: for each function $f \in \mathbb{R}^\mathbb{R}$ with $f(0) = 0$ there is some $F \in V^V$ such that $\langle Fx - Fy, Fx - Fy\rangle = f( \langle x - y, x - y\rangle )$ ($x,y \in V$)?

I'll bet no Krein space can do this.

Somewhat vaguely, the (generally nonlinear) mapping $F$ may be viewed as an "internal realization" of the function $f$.

P.S. By "non-trivial" I mean $V \ne {0}$.

P.P.S. $A^B$ means $\lbrace u | u: B \to A \rbrace$, so that $V^2$ is the Cartesian square (of $V$). Just to avoid any confusion (I hope).

• If I may, it would also be really nice if you've used LaTeX. I guess I still have 1,300 rep points to go before I can fix it myself. – Alon Amit Dec 14 '09 at 8:09
• Remark: this question does not display well in chrome on a mac (the symmetric bilinar pairing gives a huge horizontal overflow). I tried to edit it to fix it but failed. – Kevin Buzzard Dec 23 '09 at 14:02
• This sort of overflow happens in any browser, I think. Perhaps it's a limitation of jsMath? The only way I have been able to avoid them is to display the offending equation or else reword the sentence... – José Figueroa-O'Farrill Dec 24 '09 at 4:37
• Works fine for me in firefox. Maybe this discussions could be moved somewhere else. Apologies for starting it. I just wanted to explain why I edited. – Kevin Buzzard Dec 25 '09 at 12:01

## 1 Answer

I think that it is possible with a large enough vector space $V$. I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not. The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.

### Version 1

I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $x \in V_n \setminus V_{n-1}$, let $[x]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(x) = [x]$. Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

### Version 2

Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$. In this case, let $W_n$ be the vector space of formal linear combinations of elements of $(V_n \setminus V_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V_{n+1} = V_n \oplus W_n$. In this case, $W_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$. For any fixed $f$, define $F(x) = [x,f]$.

As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted. Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$. Finally, as before, $$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$ with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.

### Version 3

Ady reminds me that the second version still misses the condition that the bilinear form on $V$ should be non-degenerate. I think that the same trick works a third time: We can just enlarge $V$ to also guarantee this condition. This time let $W_n$ be as in the second version, and let $$V_{n+1} = V_n \oplus W_n \oplus V_n^*,$$ where $V_n^*$ is the (algebraic) dual vector space to $V_n$. Define the bilinear form on $V_n \oplus W_n$ as in version 2, and define $F$ as in version 2. The bilinear form on $V_n^*$ is unrestricted, and so is the bilinear pairing between $V_n^*$ and $W_n$. Finally the bilinear pairing between $V_n^*$ and $V_n$ should be the canonical pairing $\langle \phi, x \rangle = \phi(x)$. This guarantees that for every vector $x \in V_n$, there exists $y \in V_{n+1}$ such that $\langle y,x \rangle = 1$.

Every version of the construction is cheap in the sense that the image of $F$ is a linearly independent set. Moreover, in the second and third versions, the image of $F$ is far from a basis. My feeling is that it is difficult to ask for much better than that in a universal construction.

• Please correct me if I'm wrong, but this sounds to be merely a "local" construction, i.e. V = V(f). It's of course promising, but simply embedding all the V(f)'s as orthocomplemented subspaces ("blocks") of a huge space V would not be enough, I think [at first glance]. I mean, the "ultimate" space should be independent of any given function f. – Ady Dec 24 '09 at 4:45
• Yes, you are right, the construction answers a weaker question. – Greg Kuperberg Dec 24 '09 at 5:31
• I think I patched it? – Greg Kuperberg Dec 24 '09 at 5:57
• Hmm... Firstly, is it non-degenerate ? – Ady Dec 26 '09 at 1:34
• I think I patched it again? :-) – Greg Kuperberg Dec 26 '09 at 3:25