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YCor
YCor
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Let$\DeclareMathOperator\QEnd{QEnd}$Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by :

$$d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1})$$ $$+ \;(-1)^{n+1} f(a_1,...,a_{n})$$

This différentialdifferential is null for additive functions. LetsLet's call $QEnd(Z)\subset C^1$$\QEnd(Z)\subset C^1$ the set of functions whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the field of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = QEnd(Z)/B^1 = \mathbb{R}$$$$ H^1(\bar{C^n}) = \QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about Group Cohomologygroup cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by :

$$d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1})$$ $$+ \;(-1)^{n+1} f(a_1,...,a_{n})$$

This différential is null for additive functions. Lets call $QEnd(Z)\subset C^1$ the set of functions whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the field of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about Group Cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

$\DeclareMathOperator\QEnd{QEnd}$Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by :

$$d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1})$$ $$+ \;(-1)^{n+1} f(a_1,...,a_{n})$$

This differential is null for additive functions. Let's call $\QEnd(Z)\subset C^1$ the set of functions whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the field of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = \QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about group cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

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ychemama
ychemama
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Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by : $$ d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1}) + (-1)^{n+1} f(a_1,...,a_{n})$$

$$d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1})$$ This$$+ \;(-1)^{n+1} f(a_1,...,a_{n})$$

This différential is null for additive functions. Lets call $QEnd(Z)\subset C^1$ the set of functionfunctions whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the fiedlfield of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about Group Cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by : $$ d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1}) + (-1)^{n+1} f(a_1,...,a_{n})$$ This différential is null for additive functions. Lets call $QEnd(Z)\subset C^1$ the set of function whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the fiedl of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about Group Cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by :

$$d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1})$$ $$+ \;(-1)^{n+1} f(a_1,...,a_{n})$$

This différential is null for additive functions. Lets call $QEnd(Z)\subset C^1$ the set of functions whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the field of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about Group Cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

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ychemama
ychemama
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What is this cochain complex about, whose $H^1 = \mathbb{R}$?

Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by : $$ d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1}) + (-1)^{n+1} f(a_1,...,a_{n})$$ This différential is null for additive functions. Lets call $QEnd(Z)\subset C^1$ the set of function whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the fiedl of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about Group Cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?