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Ben Webster
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Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)$M \mapsto M/\langle rm-mr|r\in A\rangle$).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A$\to A \otimes A \otimes A \to A \otimes A$ but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t]$k[t] \otimes k[t] \to k[t] \otimes k[t]$ with the map given by 1 (x) t - t (x) 1$1 \otimes t - t \otimes 1$, so when you apply coinvariants, you get two copies of k[t]$k[t]$ with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of k[t] (x) k[t]$k[t] \otimes k[t]$ is 2 gets you halfway there.

EDIT AGAIN: Actually, any Koszul algebra has its Hochschild homology bounded above by its global dimension. This is clear from the existence of the diagonal Koszul resolution.

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of k[t] (x) k[t] is 2 gets you halfway there.

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants ($M \mapsto M/\langle rm-mr|r\in A\rangle$).

Your definition used the "bar-complex" resolution of the form $\to A \otimes A \otimes A \to A \otimes A$ but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form $k[t] \otimes k[t] \to k[t] \otimes k[t]$ with the map given by $1 \otimes t - t \otimes 1$, so when you apply coinvariants, you get two copies of $k[t]$ with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of $k[t] \otimes k[t]$ is 2 gets you halfway there.

EDIT AGAIN: Actually, any Koszul algebra has its Hochschild homology bounded above by its global dimension. This is clear from the existence of the diagonal Koszul resolution.

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Ben Webster
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Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of k[t] (x) k[t] is 2 gets you halfway there.

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of k[t] (x) k[t] is 2 gets you halfway there.

added 231 characters in body
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Ben Webster
  • 44.7k
  • 12
  • 126
  • 260

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

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Ben Webster
  • 44.7k
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  • 260
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