Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:54:39Z http://mathoverflow.net/feeds/question/72791 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72791/is-the-endomorphism-algebra-of-a-dualizable-bimodule-necessarily-finite-dimension Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional? André Henriques 2011-08-12T20:07:45Z 2011-08-17T23:58:27Z <p>Let $k$ be field. Let $A$, $B$ be $k$-algebras, and let ${}_AM_B$ be a dualizable bimodule.</p> <p><i><b>Pre-Question</b> (too naive)<b>:</b></i> Is the algebra of $A$-$B$-bilinear endomorphisms of $M$ necessarily finite dimensional?<br> <i><b>Answer</b>: No.</i> Take $A$ some infinite dimensional commutative algebra, and $M={}_AA_A$. Then $End({}_AA_A)=A$ is not finite dimensional.</p> <blockquote> <p><i><b>Question</b>:</i> Assume that $A$ and $B$ have finite dimensional centers. Is it then true that the algebra of $A$-$B$-bilinear endomorphisms of $M$ has to be finite dimensional?<br></p> </blockquote> <p><i>Special case for which I know the answer to be positive:<br></i> If $k=\mathbb C$ or $\mathbb R$, and if we're in a <i>C*</i>-algebra context, then I know how to prove that $End({}_AM_B)$ is finite dimensional. But my proof relies on certain inequalities, and it does not generalize.</p> <hr> <p><b><i>Definitions:</i></b><br> A bimodule ${}_AM_B$ is called <i>left dualizable</i> if there is an other bimodule ${}_BN_A$ (the left dual) and maps $r:{}_AA_A\to {}_AM\otimes_BN_A$ and $s:{}_BN\otimes_AM_B\to {}_BB_B$ such that $(1\otimes s)\circ(r\otimes 1) = 1_M$ and $(s\otimes 1)\circ(1\otimes r) = 1_N$.</p> <p>There's a similar definition of right dualizability. <s>I'm guessing that right dualizability is not equivalent to left dualizability, but I don't know a concrete example that illustrates the difference between these two notions.</s></p> <p>In my question above, I've just used the term "dualizable", which you should interpret as "both left and right dualizable".<br><br> There's also the notion of <i>fully dualizable</i>, which means that the left dual has its own left dual, which should in turn have its own left dual etc., and similarly for right duals. Once again, I'm a bit vague as to whether all these infinitely many conditions are really needed, or whether they are implied by finitely many of them.</p> <p><i>PS:</i> I hope that I didn't mix my left and my right.</p> http://mathoverflow.net/questions/72791/is-the-endomorphism-algebra-of-a-dualizable-bimodule-necessarily-finite-dimension/72873#72873 Answer by Kevin Walker for Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional? Kevin Walker 2011-08-14T16:40:15Z 2011-08-14T18:40:31Z <p>This doesn't directly answer your question, but I think that it is relevant. Also, it's ridiculously long, but hopefully that will make for easier reading than a shorter, vaguer version.</p> <p>Let's think of the center of an algebra $A$ and the space of bimodule maps from $_AA_A$ to itself. Pictorially, we think of this vector space as associated to a circle subdivided into an incoming and outgoing interval and labeled entirely by $A$. (Later we will consider circles labeled by various combinations of $A$, $B$, $M$, and $N$.)</p> <p>We should (in TQFTish contexts) think of the finite dimensionality of a vector space $V$ as meaning there is another vector space $W$, together with maps $r:k\to V\otimes W$ and $s: V\otimes W \to k$, which satisfy the zig-zag identity. (See the definition of dualizable in André's original question.)</p> <p>But now we should smell a rat. If the center of $A$ is associated to an (appropriately decorated) circle, then the maps $r$ and $s$ should be associated to annuli (a circle cross a 1-dimensional cup or cap, like the plumbing under a kitchen sink). <img src="http://kitchensink-s.com/wp-content/uploads/2011/04/kitchen-sink-plumbing1-200x300.jpg" alt="alt text"> The "rat" is nonlocality; we are in danger of violating the dogma that everything should be local. Circles and annuli might seem simple, but intervals and disks are simpler still. We should try to replace the condition that $A$ has finite dimensional center with a local criterion.</p> <p>This can done as follows. Roughly speaking, we want to construct a fully extended 1+1-dimensional TQFT-like structure. The 0- and 1-dimensional parts can built for any algebra (or linear 1-cateogy) $A$. The existence of the 2-dimensional part, generated by (2-dimensional) cups, caps and saddles which satisfy relations corresponding to Morse cancellations, will imply that $A$ has finite dimensional center.</p> <p>Now for the details. To a point we associate $A$ (or the category of represenations of $A$, if we prefer). </p> <p>Our 1-manfolds will be split into "incoming" and "outgoing" parts. The boundary points of the 1-manifolds will be either left-facing or right-facing, and these will correspond to left and right actions of $A$.</p> <ul> <li><p>To an "outgoing" interval we associate the bimodule $_AA_A$.</p></li> <li><p>To an "incoming" interval associate the linear dual bimodule $_A(A^*)_A$.</p></li> <li><p>To an interval subdivided into incoming and outgoing halves, and with the endpoints facing left, we associate the bimodule $Hom_A(A\to A)$. This is isomorphisc as a bimodule to $_AA_A$ via the correspondence $a \mapsto (x \mapsto ax)$.</p></li> <li><p>To an interval subdivided into incoming and outgoing halves, and with the endpoints facing right, we associate the bimodule $_AHom(A\to A)$. This is isomorphisc as a bimodule to $_AA_A$ via the correspondence $a \mapsto (x \mapsto xa)$.</p></li> <li><p>To a disjoint union of an incoming and outgoing interval we associate $Hom_k(A \to A)$, thought of as a quadramodule. (The action of $a\otimes b\otimes c\otimes d$ on the function $x\mapsto f(x)$ is $x\mapsto af(cxd)b$.)</p></li> </ul> <p>We re allowed to glue intervals together if left-facing joins to right-facing and incoming [outgoing] joins to incoming [outgoing]. Gluing outgoing intervals corresponds to taking tensor product over $A$. Gluing incoming intervals corresponds to taking cotensor product over $A$. It follows that</p> <ul> <li><p>To a circle composed of an incoming and an outgoing interval (a "bigon"), we associate the space of bimodule maps from $_AA_A$ to itself. This is just the center of $A$ via the correspondence $z \mapsto (x \mapsto zx = xz)$.</p></li> <li><p>To a circle that is entirely outgoing we associate the "coinvariants" $A/\langle xy \sim yx\rangle$.</p></li> <li><p>To a circle that is entirely incoming we associate the space of traces on $A$ -- $f\in A^*$ such that $f(xy) = f(yx)$.</p></li> </ul> <p>We will only need to consider the first sort of circle above (the bigon).</p> <p>Now for the 2-dimensional part. Rather than describe the full structure one might want, I'll concentrate on the parts of the 2d structure which are relevant to the question about centers. The generators for the 2d part correspond to Morse critical points (index 0 (cup), index 1 (saddle), index 2 (cap)). These come in various flavors, depending on how the top and bottom 1-manfolds are divided into incoming and outgoing. Since we are only interested in bigon-type circles, we will only consider one flavor of cup and cap. But we will need two flavors of saddle.</p> <ul> <li><p>The cup is an element of the center $Z(A)$ of $A$. There is a canonical choice, $1\in Z(A) \subset A$.</p></li> <li><p>The cap is a function $Z(A) \to k$. This is extra data.</p></li> <li><p>The "easy" saddle is a quadramodule map from $Hom_A(A\to A) \otimes_k {}_AHom(A\to A)$ to $Hom_k(A\to A)$. There is a canonical choice for the easy saddle which sends $(x\mapsto ax)\otimes(x \mapsto xc)$ to $(x \mapsto axc)$.</p></li> <li><p>The "hard" saddle goes the other way, from $Hom_k(A\to A)$ to $Hom_A(A\to A) \otimes_k {}_AHom(A\to A)$. I don't think there is a canonical choice for this -- it's extra data.</p></li> </ul> <p>We require the above 2d data (cup, cap, easy saddle, hard saddle) to satisfy identities corresponding to Morse cancelations. </p> <p>One identity involves the cup and the easy saddle and is automatically satisfied (assuming we make the canonical choices for cup and easy saddle). More specifically we go from $Hom_A(A\to A)$ via the cup to $Hom_A(A\to A) \otimes_k Hom_{A\times A^{op}}(A\to A)$ and then via the easy saddle back to $Hom_A(A\to A)$, and this composition is the identity. (And similarly for $_AHom(A\to A)$.)</p> <p>The other identity involves the cap and the hard saddle. We go from $Hom_A(A\to A)$ via the hard saddle to $Hom_A(A\to A) \otimes_k Hom_{A\times A^{op}}(A\to A)$ and then via the cap back to $Hom_A(A\to A)$ (and similarly for $_AHom(A\to A)$). This composition is required to be the identity.</p> <p>The above identities should be thought of as 2d versions of the zig-zag identity.</p> <p><strong>Definition.</strong> An algebra $A$ has Property X is there exists a cap and hard saddle satisfying the above identities. (Recall that the cup and easy saddle come for free.)</p> <p><strong>Observation 1.</strong> If $A$ has Property X, then $A$ has finite dimensional center. Proof: Define $r:k\to Z(A)\otimes Z(A)$ to be the cup followed by the hard saddle induced map from $Hom_{A\times A^{op}}(A\to A)$ to $Hom_{A\times A^{op}}(A\to A) \otimes Hom_{A\times A^{op}}(A\to A)$. Define $s:Z(A)\otimes Z(A)\to k$ to be the similar map composed of the easy saddle followed by the cap. The 1d zig-zag identity follow from the 2d Morse cancellation identities.</p> <p>In summary, Property X is our local replacement for "finite dimensional center".</p> <p><strong>Remark.</strong> I (weakly) suspect that algbras with finite dimensional centers which do not satisfy Property X are relatively rare. If it turns out that Property X is very strong (e.g. if it implies that $A$ is finite dimensional semisimple), then I'll be a little bit embarrassed (but only weakly surprised).</p> <p>Now, finally, we get to the punch line.</p> <p><strong>Observation 2.</strong> If $A$ and $B$ have Property X and $_AM_B$ is dualizable, then $End(M)$ is finite dimensional.</p> <p><strong>Proof.</strong> The idea is to construct a 2d TQFT, similar to the one above, in which manifolds are divided into $A$-colored parts and $B$-colored parts, with the interfaces between $A$ and $B$ labeled by $M$ or $N$ as appropriate. (Recall that $N$ is the dual of $M$.) So 1-manifolds in this TQFT can be independently either incoming or outgoing, either $A$-colored or $B$-colored, and (if they are part of the boundary of a 2d bordism) either top or bottom. We will construct maps $$u: k\to End(M) \otimes End(N)$$ and $$n: End(N) \otimes End(M) \to k$$ which satisfy the zig-zag identity. It will follow that $End(M)$ is finite dimensional.</p> <p>We will need $M$ to be both right and left dualizable, with the right and left duals both $N$. Let $r_l, s_l, r_r, s_l$ be the left and right 1d cups and caps.</p> <p>The map $u$ corresponds to a bigon cross a 1d cup, half colored by $A$ and half colored by $B$. More specifically, $u$ is the composition from $k$ </p> <ul> <li><p>to $Hom_{A\times A^{op}}(A\to A)$ via the cup for $A$</p></li> <li><p>to $Hom_{A\times A^{op}}(A\to M\otimes_B N)$ via $r_l$</p></li> <li><p>to $Hom_{A\times A^{op}}(M\otimes_B N\to M\otimes_B N)$ via $s_r$</p></li> <li><p>to $Hom_{B\times A^{op}}(M\to M) \otimes Hom_{A\times B^{op}}(N\to N)$ via the hard saddle for $B$.</p></li> </ul> <p>Similarly, $n$ is the composition from $Hom_{A\times B^{op}}(N\to N) \otimes Hom_{B\times A^{op}}(M\to M)$</p> <ul> <li><p>to $Hom_{B\times B^{op}}(N\otimes_A M\to N\otimes_A M)$ via the easy saddle for $A$</p></li> <li><p>to $Hom_{B\times B^{op}}(B \to N\otimes_A M)$ via $r_r$</p></li> <li><p>to $Hom_{B\times B^{op}}(B \to B)$ via $s_l$</p></li> <li><p>to $k$ via the cap for $B$.</p></li> </ul> <p>The zig-zag identity for $u$ and $n$ follows from the right and left zig-zag identities for $M$ and the Morse cancellation identities for $A$ and $B$.</p> <p><strong>Remark.</strong> Unless I've made a mistake, we only need one of $A$ or $B$ to have Property X.</p> <p><strong>Remark.</strong> If we reverse the roles of $A$ and $B$ above, we get different maps $u': k\to End(M) \otimes End(N)$ and $n': End(N) \otimes End(M)$. (The zig-zag for $u$ and $n$ requires that $B$ and Property X, while the zig-zag for $u'$ and $n'$ requires that $A$ have Property X.) If the 2d TQFT structure is more complete, with more flavors of cups, caps and saddles, then one can show that $u=u'$ and $n=n'$.</p> <p>(This would be all be much easier with pictures. I apologize for the lack of them.)</p> http://mathoverflow.net/questions/72791/is-the-endomorphism-algebra-of-a-dualizable-bimodule-necessarily-finite-dimension/73110#73110 Answer by Evan Jenkins for Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional? Evan Jenkins 2011-08-17T23:58:27Z 2011-08-17T23:58:27Z <p>This is more or less orthogonal to Kevin's answer; he puts a stronger restriction on $A$ and $B$, while I put a stronger restriction (perhaps too strong!) on ${_A}M_B$. I likewise apologize for the lack of pictures.</p> <p>If we assume that</p> <ol> <li>${_A}M_B$ has a 2-sided dual ${_B}N_A$, and</li> <li>the duality pairings exhibit ${_B}N \otimes_A M_B$ as a retract of ${_B}B_B$,</li> </ol> <p>then $\operatorname{End}({_A}M_B)$ is dualizable as a $Z(A)$-module, and hence, if $Z(A)$ is finite dimensional, it will follow that $\operatorname{End}({_A}M_B)$ is finite dimensional.</p> <p>To prove this, we exhibit a duality pairing between ${_{Z(A)}}\operatorname{End}({_A}M_B)$ and $\operatorname{End}({_B}N_A)_{Z(A)}$.</p> <p>The evaluation map $\epsilon: {_{Z(A)}}\operatorname{End}({_A}M_B) \otimes_k \operatorname{End}({_B}N_A)_{Z(A)} \to {_{Z(A)}}Z(A)_{Z(A)}$ is given by first mapping to $\operatorname{End}({_A}M \otimes_B N_A)$ and then pre- and post-composing by the appropriate unit and counit, respectively, from the duality pairings between $M$ and $N$ to get an endomorphism of ${_A}A_A$, i.e., an element of $Z(A)$.</p> <p>The coevaluation map $\eta: k \to \operatorname{End}({_B}N_A) \otimes_{Z(A)} \operatorname{End}({_A}M_B)$ is given by identifying the codomain with $\operatorname{End}({_B}N \otimes_A M_B)$, which follows from condition 2, whence $\eta$ is just the unit of this algebra.</p> <p>That $\epsilon$ and $\eta$ give a duality pairing follows from condition 2. (The nontrivial morphism in one of the triangle diagrams sends a map $f: {_A}M_B \to {_A}M_B$ to the identity on ${_A}M_B$ tensored with the "trace" of $f$ via the duality pairings. Condition 2 tells us that the identity string can absorb the trace bubble, and similarly for the other triangle diagram.)</p> <p>Unfortunately, condition 2 (or its "mirror," which would show dualizability as a $Z(B)$-module) seems very strong, but it is crucial for the above proof.</p>