On The GICAR Algebra and The Hausdorff Moment Problem

GICAR Algebra and Its Ordered Group

Definition (GICAR algebra). The GICAR algebra is the AF algebra described by the following ($C^*$-algebra inductive limit) Bratteli diagram–Pascal’s triangle. https://q.uiver.app/?q=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

Let’s first compute its $K_0$ group.

Theorem ($K_0$ of GICAR). Continuity of $K_0$ (Theorem 6.3.2 in rørdam_larsen_laustsen_2010) means the $K_0$ group of the GICAR algebra is described by the following (group inductive limit) Bratteli diagram. https://q.uiver.app/?q=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

Note that the edges represent the group homomorphism taking $1\mapsto 1.$ This group is $\mathbb{Z}[x]$.

Proof. $K_0$ of the GICAR algebra is isomorphic to the group generated by the following Bratteli diagram (*). https://q.uiver.app/?q=WzAsOSxbMiwwLCJcXGxhbmdsZSAxIFxccmFuZ2xlIl0sWzEsMSwiXFxsYW5nbGUgeCBcXHJhbmdsZSJdLFszLDEsIlxcbGFuZ2xlIDEgLSB4IFxccmFuZ2xlIl0sWzAsMiwiXFxsYW5nbGUgeF4yIFxccmFuZ2xlIl0sWzIsMiwiXFxsYW5nbGUgeCgxIC0geCkgXFxyYW5nbGUiXSxbNCwyLCJcXGxhbmdsZSAoMSAtIHgpXjIgXFxyYW5nbGUiXSxbMCwzLCJcXHZkb3RzIl0sWzIsMywiXFx2ZG90cyJdLFs0LDMsIlxcdmRvdHMiXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMiwiIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsxLDMsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSw0LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsNCwiIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDUsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0= where each entry is a subgroup of $\mathbb{Z}[x].$ For proof, consider the following diagram, where the red arrows map generator to generator, where the generator for $\mathbb{Z}$ is $1$. https://q.uiver.app/?q=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 Observe that in each triangle of (*), we have $A = B + C$ (**). https://q.uiver.app/?q=WzAsNixbMSwxLCJcXGxhbmdsZSBBIFxccmFuZ2xlIl0sWzAsMiwiXFxsYW5nbGUgQlxccmFuZ2xlIl0sWzIsMiwiXFxsYW5nbGUgQyBcXHJhbmdsZSJdLFswLDMsIlxcdmRvdHMiXSxbMiwzLCJcXHZkb3RzIl0sWzEsMCwiXFx2ZG90cyJdLFswLDEsIiIsMSx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMCwyLCIiLDEseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d Therefore, we can treat the group direct sums as simply addition: ie. we can treat $\langle x^2\rangle\oplus \langle x(1 - x)\rangle\oplus \langle (1 - x)^2\rangle$ as $\langle a_1x^2 + a_2x(1 - x) + a_3(1 - x)^2\ |\ a_i\in \mathbb{Z}\rangle.$ Moreover, (**) means that $\langle a_1x^n + a_2x^{n - 1}(1 - x) + … + a_n(1 - x)^n\ |\ a_i\in \mathbb{Z}\rangle = \mathbb{Z}[x]_{\le n}$ (degree $\le$ $n$) because $1, x, …, x^n$ belong to this subgroup and the max degree polynomial in the subgroup is degree $n.$ Thus, the $K_0$ group we are computing is the inductive limit https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXG1hdGhiYntafVt4XV97XFxsZSAwfSJdLFsyLDAsIlxcbWF0aGJie1p9W3hdX3tcXGxlIDF9Il0sWzQsMCwiXFxtYXRoYmJ7Wn1beF1fe1xcbGUgMn0iXSxbNiwwLCJcXGNkb3RzIl0sWzAsMSwiXFxpb3RhXzAiXSxbMSwyLCJcXGlvdGFfMSJdLFsyLDMsIlxcaW90YV8yIl1d where the connecting maps are inclusions by (**). Hence, our group must be $\mathbb{Z}[x].$∎

Now let’s take a look at the ordered $K_0$ group. We first present a result of Halperin in Czarnecki1964.

Lemma (Halperin’s Lemma). Let $p(x)\in \mathbb{R}[x]$ be non-negative on the interval $[a, b].$ Then, $p(x)$ is the sum of the following forms: $(x - a)q(x)^2,$ $(b - x)q(x)^2,$ $q(x)^2$ where $q(x)\in \mathbb{R}[x].$

Proof. We proceed by induction on the degree of $p(x).$ If $p(x) = c,$ then we can write $c = (\sqrt{c})^2.$ Now suppose $p(x)$ is a degree $n\ge 1$ polynomial. Taking $m = \min_{x\in [a, b]} p(x),$ we have $p(x) = p(x) - m + (\sqrt{m})^2,$ so it suffices to prove the hypothesis for $p(x) - m.$ Since $m = \min_{x\in [a, b]} p(x),$ we can factor $p(x) - m = (x - r)q(x)$ for some $r\in [a, b].$ Here we consider two cases.

1. If $r\in (a, b),$ then $q(r) = 0$ since $x - r$ changing sign as $x$ passes $r$ forces $q(x)$ to change sign with it. Hence, $p(x) - m = (x - r)q(x) = (x - r)^2s(x).$ Since $\deg s(x) < n,$ we can invoke inductive hypothesis and are done.

2. Suppose $r = a$ or $r = b.$ Without loss of generality, suppose $r = a,$ for the $r = b$ case is symmetric. We have $p(x) - m = (x - a)q(x).$ Since $\deg q(x) < n,$ we know $q(x)$ is the sum of the forms $(x - a)s(x)^2,$ $(b - x)s(x)^2,$ $s(x)^2.$ It suffices to prove that our inductive hypothesis holds for $p(x) - m$ when $q(x) = (b - x)s(x)^2.$ Observe that

   \(\begin{aligned} (x - a)(b - x) &= (x - a)(b - x)\frac{(x - a + b - x)}{b - a}\\ &= \frac{1}{b - a}[(b - x)(x - a)^2 + (x - a)(b - x)^2]. \end{aligned}\)

   Since $\frac{1}{b - a} > 0,$ we can put $\frac{1}{\sqrt{b - a}}$ into each of the squares. Then, $p(x) - m$ satisfies our inductive hypothesis. ∎

Lemma (Basis for Positive Cone). A polynomial $p(x)\in \mathbb{Z}[x]$ is non-negative on ${0, 1}$ and strictly positive on $(0, 1)$ if and only if it can be written as \(p(x) = \sum_k a_kx^{e_k}(1 - x)^{f_k}\) where $a_k\in \mathbb{Z}$ and $a_k > 0.$

Proof. The $\impliedby$ direction is easy. Let’s consider the $\implies$ direction. Suppose $p(x)\in \mathbb{Z}[x]$ is non-negative on ${0, 1}$ and strictly positive on $(0, 1).$ We will proceed by induction on the degree of $p(x)$. It’s clear $p(x)\neq 0.$ If $p(x)$ is degree $0,$ $p(x) = c > 0.$ Note that $c = cx + c(1 - x).$ Suppose that the hypothesis holds when $\deg p(x)\le n.$ By Halperin’s Lemma, we are done. ∎

Remark. One can see that we can swap $\mathbb{Z}$ with $\mathbb{R}$ without any problems.

Theorem (Ordered $K_0$ of GICAR). The ordered $K_0$ group of GICAR is $\mathbb{Z}[x],$ where $\mathbb{Z}[x]^+$ consists of $0$ and elements of $\mathbb{Z}[x]$ that are non-negative on ${0, 1}$ and strictly positive on $(0, 1).$

Proof. A direct consequence of continuity of $K_0$ (Theorem 6.3.2 in rørdam_larsen_laustsen_2010) and the Basis for Positive Cone Lemma. ∎

Hausdorff Moment Problem

The Hausdorff moment problem asks for necessary and sufficient conditions for a sequence $a_0, a_1, …$ to satisfy \(a_k = \int_0^1 x^k\ d\mu(x)\) for some positive measure $\mu.$ We will henceforth suppress the integration bounds.

Theorem (Necessary condition). Let $a_0, a_1, …$ satisfy \(a_k = \int x^k\ d\mu(x)\) for some positive measure $\mu.$ Then, the Pascal triangle generated by writing $a_0, a_1, …$ along the left diagonal must have non-negative entries, ie. every entry in the following diagram must be non-negative. https://q.uiver.app/?q=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&macro_url=%5Cdef%5Crddots%231%7B%5Ccdot%5E%7B%5Ccdot%5E%7B%5Ccdot%5E%7B%231%7D%7D%7D%7D

Proof. Integral of positive function with respect to positive measure is positive. Thus, our condition is indeed a necessary condition because by replacing the values in the Pascal triangle with their integral counterparts, we get the following diagram. https://q.uiver.app/?q=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&macro_url=%5Cdef%5Crddots%231%7B%5Ccdot%5E%7B%5Ccdot%5E%7B%5Ccdot%5E%7B%231%7D%7D%7D%7D ∎

Lemma (Approximation by Positive Polynomials). Functions $f\in C([0, 1])$ such that $f\ge 0$ can be uniformly approximated by polynomials that are strictly positive on $[0, 1].$

Proof. Let $f\in C([0, 1])$ where $f\ge 0$ and $\epsilon > 0$ be arbitrary. Observe that $\sqrt{f}\in C([0, 1])$ so for any $\delta > 0,$ by the Weierstrass theorem, there exist a polynomial $p$ such that $\sup_{x\in [0, 1]} |\sqrt{f(x)} - p(x)| < \delta.$ Let $M = \sup_{x\in [0, 1]} |\sqrt{f(x)}|.$ We have

This means that there exists a polynomial $q\ge 0$ such that $\sup_{x\in [0, 1]} |f(x) - q(x)| < \frac{\epsilon}{2}.$ Then, take $q(x) + \frac{\epsilon}{2}$ to be our strictly positive approximating polynomial: $\sup_{x\in [0, 1]} |f(x) - (q(x) + \frac{\epsilon}{2})| < \epsilon.$ ∎

Theorem. The necessary condition is also a sufficient condition.

Proof. Suppose the sequence $a_0, a_1, …$ satisfies the necessary condition. By the Weierstrass theorem we can define a bounded (***) linear functional $\psi: C([0, 1])\rightarrow \mathbb{R}$ by setting $\psi(x^k) = a_k.$ We claim that $\psi$ is a positive linear functional. Let $f\in C([0, 1])$ be positive. The Approximation Lemma tells us that there exist a sequence of polynomials $p_n,$ each non-negative on ${0, 1}$ and strictly positive on $(0, 1),$ that converge uniformly to $f.$ Because $a_1, a_2, …$ satisfies the necessary condition, we know $\psi(x^e(1 - x)^f)\ge 0$ for all valid $e, f.$ Then, by the Basis for Positive Cone Lemma and continuity of $\psi,$ we know that $\psi(f)\ge 0.$ Finally, we can conclude by using the Riesz-Markov-Kakutani representation theorem that there exists a Radon (hence positive) measure $\mu$ such that for all $f\in C([0, 1])$ \(\psi(f) = \int f(x)\ d\mu(x).\) We are done. ∎

Remark. (***) Is actually circular logic. We leave the fix as an exercise (:p).

References

• Czarnecki, A., Federico, P. J., Van Vleck, F. S., Halperin, I., Parker, F. D., Day, G. W., Schubert, S. R., & Younglove, J. N. (1964). Mathematical Notes. The American Mathematical Monthly, 71(4), 403–413. https://doi.org/10.1080/00029890.1964.11992254
• Dixmier, J. (1967). On some C-algebras considered by Glimm. *Journal of Functional Analysis, 1(2), 182–203.
• Dixmier, J. (1977). C-algebras*. North-Holland.
• Glimm, J. G. (1960). On a certain class of operator algebras. Transactions of the American Mathematical Society, 95(2), 318–340.
• Rørdam, M., Larsen, F., & Laustsen, N. (2010). An introduction to K-theory for $C^$-algebras*. Cambridge University Press.
• Wikimedia Foundation. (2023). Binomial coefficient. Wikipedia. Retrieved from https://en.wikipedia.org/wiki/Binomial_coefficient