# Tagged Questions

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### Hochschild cohomology of skew polynomial rings

Definition The skew polynomial algebra over $\mathbb{C}$ is defined as $\mathbb{C}\langle x,y\rangle/(xy-yx+x)$ or alternatively as $\mathbb{C}[x,y,\sigma]$, where $\sigma$ is the automorphism on ...
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### Global dimension of translation algebra

What is the Hochschild cohomological dimension of the "translation algebra": $\mathbb{C}\langle x,y\rangle/(xy-yx-x)$? I expect it to be $2$, but I haven;t found a serious argument as to why this ...
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### The first Weyl algebra is Calabi-Yau

Why is the Weyl algebra $A_1(k)$ over a field $k$ Calabi-Yau? (My definition of Calabi-Yau is Ginzburg's)
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### Example of excision in Hochschild homology

The excision theorem for Hochschild homology introduced by Wodzicki seems like a very powerful tool (as scision was hyper-useful in topology). However, I cannot actually seem to think of a result ...
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### Coproducts and Hochschild

I $\{X_i\}$ is a small family of associative $\mathbb{C}$-algebras and $X$ is their free product. Then I have two questions: 1) Why is $X$ their coproduct? 2) Is the Hochschild homology of X ...
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### Monoidal categories and tensor products

Does the multiplication $-\square-$ biendofunctor in a Monoidal category, $\mathfrak{C}$ necessarily commute with coproducts? This is true in some familiar categories, such as $_RMod$, $Grp$ $CRings$ ...
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### Smooth algebras and quasi-freeness

Let A be a unital associative algebra over a field k. Then A is smooth if and only if X:=Spec(A) is smooth. That is $\Omega_{X|Spec(k)}$ is locally-free. The later module is isomorphic to ...
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### Reference for proof of homotopy invarance of Cyclic cohomolgy

I'm looking for a good reference for a proof of the homotopy invariance of cyclic (co)homology. I'm following a refernce book by Joachim Cuntz, the proofs are ommited therein, or only shown in the ...
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### A question about C* probability spaces.

If I have $C(\Omega)$: the commutative C*-Algebra of all continuous random variables then the C*Algebra $C(\Omega)$ acts on the Hilbert space $L^2(\Omega)$ by left multiplication, and so we have the ...