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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeMar 23rd 2014
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   Author: Mike Shulman
   Format: MarkdownItexI am making a mistake somewhere, can you help me find it? The monoidal structure on (unbounded) chain complexes admits two symmetries, i.e. there are two symmetric monoidal structures with the same underlying monoidal structure. The "usual" symmetry inserts a sign according to degree, $x\otimes y \mapsto (-1)^{|x| |y|} y\otimes x$, while the "wrong" symmetry has no sign, $x \otimes y\mapsto y\otimes x$. Now the condition to be a symmetric monoidal model category says nothing about the symmetry, so it seems that both of these should be symmetric monoidal model categories (with the projective model structure) and hence give rise to monoidal structures on the $(\infty,1)$-category presented by unbounded chain complexes. But it's a general fact in any stable monoidal $(\infty,1)$-category that the symmetry *always* introduces a sign, as expressed for instance in May's axiom TC1 for a monoidal triangulated category. What's wrong?

   I am making a mistake somewhere, can you help me find it? The monoidal structure on (unbounded) chain complexes admits two symmetries, i.e. there are two symmetric monoidal structures with the same underlying monoidal structure. The “usual” symmetry inserts a sign according to degree, $x\otimes y \mapsto (-1)^{|x| |y|} y\otimes x$, while the “wrong” symmetry has no sign, $x \otimes y\mapsto y\otimes x$. Now the condition to be a symmetric monoidal model category says nothing about the symmetry, so it seems that both of these should be symmetric monoidal model categories (with the projective model structure) and hence give rise to monoidal structures on the $(\infty,1)$-category presented by unbounded chain complexes. But it’s a general fact in any stable monoidal $(\infty,1)$-category that the symmetry always introduces a sign, as expressed for instance in May’s axiom TC1 for a monoidal triangulated category. What’s wrong?

2. • CommentRowNumber2.
   • CommentAuthorZhen Lin
   • CommentTimeMar 23rd 2014
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   Author: Zhen Lin
   Format: MarkdownItexIsn't there also a sign rule in the definition of $\otimes$ itself?

   Isn’t there also a sign rule in the definition of $\otimes$ itself?

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeMar 23rd 2014
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   Author: Mike Shulman
   Format: MarkdownItexAh, yes... are you saying that that prevents the wrong symmetry from even existing? That may be it.

   Ah, yes… are you saying that that prevents the wrong symmetry from even existing? That may be it.

4. • CommentRowNumber4.
   • CommentAuthorZhen Lin
   • CommentTimeMar 23rd 2014
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   Author: Zhen Lin
   Format: MarkdownItexThat was what I was thinking, yes. The "wrong" symmetry fails to be a morphism at all: for $(x, y)$ of degree $(1, 1)$, we have $d (x \otimes y) = d x \otimes y - x \otimes d y$, so compatibility with differentials forces a negative sign.

   That was what I was thinking, yes. The “wrong” symmetry fails to be a morphism at all: for $(x, y)$ of degree $(1, 1)$, we have $d (x \otimes y) = d x \otimes y - x \otimes d y$, so compatibility with differentials forces a negative sign.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMar 24th 2014
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   Author: Mike Shulman
   Format: MarkdownItexThanks! I guess I was mixing up chain complexes and graded objects.

   Thanks! I guess I was mixing up chain complexes and graded objects.