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YCor
YCor
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Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq Hom(A, \mathbb{Z})$$H^{1}(A, \mathbb{Z}) \simeq \operatorname{Hom}(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?

Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq Hom(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?

Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq \operatorname{Hom}(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?

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Piotr Pstrągowski
Piotr Pstrągowski
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Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq Hom(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?

Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq Hom(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?

Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers. We will consider this as a discrete abelian group.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq Hom(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?

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Piotr Pstrągowski
Piotr Pstrągowski
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Cohomology of the Baer-Specker group

Let $A = \prod_{i \in \mathbb{N}} \mathbb{Z}$ be the Baer-Specker group; that is, a countably infinite product of the integers.

Are the higher cohomology groups $H^{*}(A, \mathbb{Z})$ known?

In the first degree, we have $H^{1}(A, \mathbb{Z}) \simeq Hom(A, \mathbb{Z})$, which is known to be a countably infinite direct sum of $\mathbb{Z}$. Is $H^{*}(A, \mathbb{Z})$ isomorphic to an exterior algebra over $H^{1}(A, \mathbb{Z})$, like we would expect from a finite product?