## A Resolution of a Tensor Algebra

This post is about projective resolutions of algebras, thought of as a bimodules over themselves.  As long as $B$ is an associative, unital algebra (which it always will be in this post), there is a canonical projective resolution of $B$, called the bar resolution, which is sufficient for most purposes.  However, this resolution is of infinite length, and so it isn’t useful in bounding projective dimensions of modules.  For those purposes, it is natural to look for finite projective resolutions of $B$.

I came across such a problem in my research, and came up with limited and ultimately unhelpful results.  My interest was in the case that $B$ is a tensor algebra $T_AM$ of an algebra $A$ over a bimodule $M$.  Under what conditions on $A$ and $M$ would a nice, finite resolution of $T_AM$ exist?  My result is as follows:

Let $A$ be an algebra, $M$ be a bimodule over $A$, and let $\Omega^1A$ denote the kernel of the multiplication map $m: A\otimes A \rightarrow A$.  If $\Omega^1A$ is projective as an $A$ bimodule, and $Tor_1^A(M_A,_AM)=0$, then there is a projective resolution of $T_AM$ of length 3.

This is kinda neat, but its not super useful unless it can be used to produce projective resolutions of $T_AM$ modules.  Hence, the second result:

Let $A$, $M$, and $\Omega^1A$ be defined as above.  If $\Omega^1A$ is projective and $M$ is flat as a right $A$ module, then any left $T_AM$ module has a projective resolution of length 3.

Sadly, I wanted the assumptions of the former to prove the latter, which my techniques don’t.

I should mention that the assumptions I place on $A$ and $M$ are very standard conditions.  If $A$ has the property that $\Omega^1A$ is projective, $A$ is called quasi-free, or (more clearly) formally smooth

This name comes from the fact that the module $\Omega^1A$ is the module of ‘noncommutative 1-forms’.  To see this, notice that $Hom(\Omega^1A,N)=Der(A,N)$ for any $A$ bimodule $N$, and so $\Omega^1A$ is ‘formally dual’ to derivations/vector fields.  Therefore, $\Omega^1A$ being projective is the noncommutative version of the 1-forms on $Spec(A)$ constituting a vector bundle, and hence $Spec(A)$ being smooth.

A module $M$ such that $Tor_1^A(M,M)=0$ is called pseudoflat, since it is clearly weaker than being flat.  Such a property doesn’t have quite such an elegant geometric meaning, but it still has a straightforward homological meaning which is relevant.  Since $\Omega^1A$ has a natural embedding $\iota$ into $A\otimes A$, this induces a map which we will again call $\iota$ from $M\otimes _A \Omega^1A\otimes_A M$ to $M\otimes M$.  This map fits into a long exact sequence:

$0\rightarrow Tor_1^A(M,M)\rightarrow M\otimes_A \Omega^1A\otimes_A M\rightarrow M\otimes M \rightarrow M\otimes_A M \rightarrow 0$

Therefore, $M$ is pseudoflat iff $\iota:M\otimes_A \Omega^1A\otimes_AM\rightarrow M\otimes M$ is an inclusion.

For simplicity, and to make some formulas clearer, let $\overline{\otimes}:=\otimes_A\Omega^1\otimes_A$.  Note that there is always a map $\iota: N\overline{\otimes}N'\rightarrow N\otimes N'$, and that its an inclusion iff $Tor_1^A(N,N')=0$.

Consider the following complex:

$\begin{array}{c} 0 \\ \downarrow \\ T_AM\overline{\otimes} M \overline{\otimes} T_AM \\ f_2\downarrow \\ (T_AM\overline{\otimes}M\otimes T_AM) \oplus (T_AM\otimes M \overline{\otimes} T_AM) \\ f_1\downarrow \\ (T_AM\overline{\otimes} T_AM) \oplus (T_AM\otimes M \otimes T_AM) \\ f_0\downarrow \\ T_AM\otimes T_AM \\ b\downarrow \\ T_AM \end{array}$

$f_0$ acts as $\iota$ on $T_AM\overline{\otimes}T_AM$, and sends $m\in M$ to $m\otimes 1-1\otimes m$.

$f_1$ on the first summand is equal to the sum of $\iota$ and the multiplication map from $M\otimes T_AM$ to $T_AM$, and acts similarly on the second summand.

$f_2$ is $\iota\overline{\otimes} Id_{T_AM}-Id_{T_AM}\overline{\otimes} \iota$.

For arbitrary $A$ and $M$, this is a complex.  If $A$ is quasi-free, this is a projective complex.  These are both straight-forward computations.  If $M$ is pseudoflat, this complex is exact, but this is not as straight-forward.

So where does this resolution come from?  It was constructed in the most naive possible way.  The first step of the resolution pretty much has to be $T_AM\otimes T_AM$ with the multiplication map to $T_AM$.  Thus, the rest of the resolution is a resolution of the kernel of this map, $\Omega^1(T_AM)$.

So then what is a set of generators of $\Omega^1(T_AM)$?  It isn’t hard to see that it is generated by elements of the form $dt:=t\otimes 1-1\otimes t$, and that these elements satisfy the relation $d(t\tau)=d(t)\tau+td(\tau)$.  Therefore, we can generate $\Omega^1(T_AM)$ by elements of the form $\omega\in \Omega^1A$ and $dm$, $m\in M$.  This is what the second step in the resolution is.

Ok, so then what are the relations between these generators?  The obvious ones are $d(am)=d(a)m+ad(m)$ and $d(ma)=d(m)a+ad(m)$ for all $a\in A$ and $m\in M$.  Its a little bit hard to see, but the third step in the resolution is the module generated by relations of this form.  The fact that the complex is exact at that place is means that the obvious relations are the only ones, which is a neat fact to know.

Finally, there is the question of relations between the previously found relations.  This is even harder to think about, but thankfully there are some simple ones, coming from the two sides one can resolve the expression $d(amb)$.  This is the module generated by the third term in the resolution, and again exactness is equivalent to the fact that these are the only relations.

So, there it is!  Its not an earth-shaking fact, but it was fun to discover.  I’m also kind of hoping that someone can give this result a better home than me, since I have discarded it for not being relevant to my uses.  I have a proof written up for this, and the resolution of a left module $N$, which is availible on request but too big to put in this post.

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### 4 Responses to “A Resolution of a Tensor Algebra”

1. Anonymous Says:

Isn’t T_A M, like all rings, projective over itself? Are you looking for a resolution in the category of A-modules or something?

2. Aaron Bergman Says:

Unless I completely misunderstand, he’s looking at the category of A-bimodules, ie, A^op \otimes A modules.

3. Greg Muller Says:

Yup, that is correct. Though, just to be clear, I’m looking for a projetive resolution of $T_AM$ in the category of $T_AM$-bimodules.

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