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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeFeb 16th 2016
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   Author: Urs
   Format: MarkdownItexI have been expanding Idea-section and References-section at _[[smash product of spectra]]_. (I suppose all technical detail should go to the respective entries for the various models of spectra). Notice that this is distinct from the entry _[[symmetric smash product of spectra]]_. I think, or thought, it makes sense to keep these separate, but I might easily be convinced otherwise.

   I have been expanding Idea-section and References-section at smash product of spectra. (I suppose all technical detail should go to the respective entries for the various models of spectra).

   Notice that this is distinct from the entry symmetric smash product of spectra. I think, or thought, it makes sense to keep these separate, but I might easily be convinced otherwise.

2. • CommentRowNumber2.
   • CommentAuthorzskoda
   • CommentTimeFeb 25th 2016
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   Author: zskoda
   Format: MarkdownItexWhat you call [[sequential spectrum]] is in written literature maybe more often called Boardman spectrum.

   What you call sequential spectrum is in written literature maybe more often called Boardman spectrum.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMar 3rd 2016
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   Author: Urs
   Format: MarkdownItexZoran, thanks, sure, I have added the terminology "Boardman spectrum" to _[[sequential spectrum]]_.

   Zoran, thanks, sure, I have added the terminology “Boardman spectrum” to sequential spectrum.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 3rd 2016
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   Author: Urs
   Format: MarkdownItexI have started to add a little section _[Properties --- Graded commutativity](https://ncatlab.org/nlab/show/smash+product+of+spectra#GradedCommutativity)_ at _[[smash product of spectra]]_ with some basic as to where it comes from and why it is the reason that plain sequential spectra do not admit a symmetric monoidal smash product.

   I have started to add a little section Properties — Graded commutativity at smash product of spectra with some basic as to where it comes from and why it is the reason that plain sequential spectra do not admit a symmetric monoidal smash product.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMar 4th 2016
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   Author: Mike Shulman
   Format: MarkdownItexI don't see how a Quillen equivalence to excisive functors can be said to explain why sequential spectra don't admit a symmetric monoidal smash product, since sequential spectra are also Quillen equivalent to plenty of categories that *do* admit a symmetric monoidal smash product.

   I don’t see how a Quillen equivalence to excisive functors can be said to explain why sequential spectra don’t admit a symmetric monoidal smash product, since sequential spectra are also Quillen equivalent to plenty of categories that do admit a symmetric monoidal smash product.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeMar 4th 2016
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   Author: Urs
   Format: MarkdownItexExcisive functors also admit a symmetric monoidal smash product. What I said (but maybe I failed expressing myself) is that the restriction of excisive functors to standard spheres with just the suspension adjuncts as maps between them (which is the right adjoint of that Quillen equivalence) breaks the symmetry.

   Excisive functors also admit a symmetric monoidal smash product.

   What I said (but maybe I failed expressing myself) is that the restriction of excisive functors to standard spheres with just the suspension adjuncts as maps between them (which is the right adjoint of that Quillen equivalence) breaks the symmetry.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeMar 4th 2016
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   Author: Mike Shulman
   Format: MarkdownItexBut what does that have to do with sequential spectra?

   But what does that have to do with sequential spectra?

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeMar 4th 2016
   • (edited Mar 4th 2016)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSequential spectra are indexed over $(\mathbb{N},+)$. The symmetry in the sum $n_1 + n_2 \to n_2 + n_1$ has no way to recognize the nontrivial symmetry in $S^{n_1}\wedge S^{n_2} \to S^{n_2} \wedge S^{n_1}$. That is the source of the issue with defining symmetric smash products of spectra, precisely the fact that this symmetry is a non-trivial automorphism of $S^{n_1+n_2}$. This is of course standard, as in example 4.1 of MMSS00 that is pointed to in the entry. What I did in the remark that you read (or maybe just skimmed?) was just to substantiate this comment by pointing out that it is indeed the Quillen map from excisive functors to sequential spectra which regards the latter as excisive functors restricted from all of $sSet^{\ast/}_{fin}$ to just the standard spheres and forgetting just those non-trivial automorphisms of spheres.

   Sequential spectra are indexed over $(\mathbb{N},+)$. The symmetry in the sum $n_1 + n_2 \to n_2 + n_1$ has no way to recognize the nontrivial symmetry in $S^{n_1}\wedge S^{n_2} \to S^{n_2} \wedge S^{n_1}$. That is the source of the issue with defining symmetric smash products of spectra, precisely the fact that this symmetry is a non-trivial automorphism of $S^{n_1+n_2}$. This is of course standard, as in example 4.1 of MMSS00 that is pointed to in the entry. What I did in the remark that you read (or maybe just skimmed?) was just to substantiate this comment by pointing out that it is indeed the Quillen map from excisive functors to sequential spectra which regards the latter as excisive functors restricted from all of $sSet^{\ast/}_{fin}$ to just the standard spheres and forgetting just those non-trivial automorphisms of spheres.

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeMar 5th 2016
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   Author: Mike Shulman
   Format: MarkdownItexWhat do you mean by "substantiate"? The comment as written says "The phenomenon in prop. 1 is the reason why there is no symmetric smash product of spectra on plain sequential spectra:" ending with a colon, which suggests to me that the subsequent paragraph about excisive functors is supposed to explain how or why prop. 1 is the reason why there is no symmetric smash product of sequential spectra, but I don't see how it does that.

   What do you mean by “substantiate”? The comment as written says “The phenomenon in prop. 1 is the reason why there is no symmetric smash product of spectra on plain sequential spectra:” ending with a colon, which suggests to me that the subsequent paragraph about excisive functors is supposed to explain how or why prop. 1 is the reason why there is no symmetric smash product of sequential spectra, but I don’t see how it does that.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeMar 5th 2016
    • (edited Mar 5th 2016)
    • PermaLink
    Author: Urs
    Format: MarkdownItexBy "substantiate" I mean that where the standard comment as in example 4.1 of MMSS00 just points out that the evident way to use indexing on $(\mathbb{N},+)$ to define spectra fails to give a symmetric product, the Quillen functor from excisive functors makes manifest that it is indeed that evident way (as opposed to any other clever idea one might come up with) which has to give the smash product: the correct smash product is Day convolution of $\infty Grpd^{\ast/}$-valued functors on $(\infty Grpd^{\ast/}_{fin}, \wedge)$ and the comparison functor is restriction of such functors along the non-full inclusion $(\mathbb{N} , +) \longrightarrow \infty Grpd^{\ast/}_{fin} : n \mapsto S^n$. Under this non-full inclusion the non-triviality of the symmetry in $(\infty Grpd_{fin}^{\ast/}, \wedge)$ is forgotten, and hence there is no way that the symmetry information in the Day convolution product structure could be retained over $(\mathbb{N},+)$

    By “substantiate” I mean that where the standard comment as in example 4.1 of MMSS00 just points out that the evident way to use indexing on $(\mathbb{N},+)$ to define spectra fails to give a symmetric product, the Quillen functor from excisive functors makes manifest that it is indeed that evident way (as opposed to any other clever idea one might come up with) which has to give the smash product: the correct smash product is Day convolution of $\infty Grpd^{\ast/}$-valued functors on $(\infty Grpd^{\ast/}_{fin}, \wedge)$ and the comparison functor is restriction of such functors along the non-full inclusion $(\mathbb{N} , +) \longrightarrow \infty Grpd^{\ast/}_{fin} : n \mapsto S^n$. Under this non-full inclusion the non-triviality of the symmetry in $(\infty Grpd_{fin}^{\ast/}, \wedge)$ is forgotten, and hence there is no way that the symmetry information in the Day convolution product structure could be retained over $(\mathbb{N},+)$

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeMar 6th 2016
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    Author: Mike Shulman
    Format: MarkdownItexAh, I see now! Thanks for explaining. Let's put that into the page.

    Ah, I see now! Thanks for explaining. Let’s put that into the page.

12. • CommentRowNumber12.
    • CommentAuthorKarol Szumiło
    • CommentTimeMar 6th 2016
    • (edited Mar 6th 2016)
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    Author: Karol Szumiło
    Format: MarkdownItexIsn't it actually even worse than that? I mean, $\mathbb{N}$ is a specific category enriched in based spaces with spheres of appropriate dimensions as mapping spaces. I don't think that this category inherits a monoidal structure from the smash product of based spaces at all (much less a symmetric one). Just to write down the action of the monoidal product on morphisms I find myself having to permute some smash coordinates and I don't see a consistent way of doing that. To be more precise, the mapping space $\mathbb{N}(m, n)$ is $S^{\wedge\{m+1, \ldots, n\}}$ (unless $m \gt n$ in which case it is a point). The inclusion into $\mathsf{Top}_*$ sends $n$ to $S^n$ and the map on morphisms $\mathbb{N}(m, n) \to \mathsf{Top}_*(S^m, S^n)$ is the adjoint of the preferred isomorphism $S^m \wedge S^{\wedge\{m+1, \ldots, n\}} \to S^n$. For $+$ to be a tensor product on $\mathbb{N}$ compatible with the smash product on $\mathsf{Top}_*$ the following square has to commute. $$ \array{ S^{\wedge\{m_0+1, \ldots, n_0\}} \wedge S^{\wedge\{m_1+1, \ldots, n_1\}} & \longrightarrow & S^{\wedge\{m_0+m_1+1, \ldots, n_0+n_1\}} \\ \downarrow & & \downarrow \\ \mathsf{Top}_*(S^{m_0}, S^{n_0}) \wedge \mathsf{Top}_*(S^{m_1}, S^{n_1}) & \longrightarrow & \mathsf{Top}_*(S^{m_0+m_1}, S^{n_0+n_1}) } $$ However, the image of $(x_0, x_1) \in S^{\wedge\{m_0+1, \ldots, n_0\}} \wedge S^{\wedge\{m_1+1, \ldots, n_1\}}$ under the composite through the lower left corner is the map $(y_0, y_1) \mapsto (y_0, x_0, y_1, x_1)$ which is not in the image of the right vertical map. Am I missing something?

    Isn’t it actually even worse than that? I mean, $\mathbb{N}$ is a specific category enriched in based spaces with spheres of appropriate dimensions as mapping spaces. I don’t think that this category inherits a monoidal structure from the smash product of based spaces at all (much less a symmetric one). Just to write down the action of the monoidal product on morphisms I find myself having to permute some smash coordinates and I don’t see a consistent way of doing that.

    To be more precise, the mapping space $\mathbb{N}(m, n)$ is $S^{\wedge\{m+1, \ldots, n\}}$ (unless $m \gt n$ in which case it is a point). The inclusion into $\mathsf{Top}_*$ sends $n$ to $S^n$ and the map on morphisms $\mathbb{N}(m, n) \to \mathsf{Top}_*(S^m, S^n)$ is the adjoint of the preferred isomorphism $S^m \wedge S^{\wedge\{m+1, \ldots, n\}} \to S^n$. For $+$ to be a tensor product on $\mathbb{N}$ compatible with the smash product on $\mathsf{Top}_*$ the following square has to commute.

    $$\array{ S^{\wedge\{m_0+1, \ldots, n_0\}} \wedge S^{\wedge\{m_1+1, \ldots, n_1\}} & \longrightarrow & S^{\wedge\{m_0+m_1+1, \ldots, n_0+n_1\}} \\ \downarrow & & \downarrow \\ \mathsf{Top}_*(S^{m_0}, S^{n_0}) \wedge \mathsf{Top}_*(S^{m_1}, S^{n_1}) & \longrightarrow & \mathsf{Top}_*(S^{m_0+m_1}, S^{n_0+n_1}) }$$

    However, the image of $(x_0, x_1) \in S^{\wedge\{m_0+1, \ldots, n_0\}} \wedge S^{\wedge\{m_1+1, \ldots, n_1\}}$ under the composite through the lower left corner is the map $(y_0, y_1) \mapsto (y_0, x_0, y_1, x_1)$ which is not in the image of the right vertical map.

    Am I missing something?

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeMar 7th 2016
    • (edited Mar 7th 2016)
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    Author: Urs
    Format: MarkdownItexKarol, that's a good point. In the spirit of MMSS00 we'd be looking at two consecutive restrictions, along the two maps $$ \mathbb{N} \longrightarrow StdSpheres \longrightarrow \infty Grpd^{\ast/}_{fin} $$ (where by $StdSpheres$ I mean $\mathbb{N}$ equipped with its enrichment by $StdSpheres(m,n) = S^{max(n-m,0)}$). While on $StdSpheres$ the structure maps of the sequential spectra are encoded by the enrichment, it would be understood that over $\mathbb{N}$ the structure maps are added in by hand using the monoidal structure. Since we need not necessarily ask that the smash product of spectra be strictly preserved along the restriction maps, only that we may somehow reconstruct it up to equivalence, maybe the failure of $StdSpheres$ to support a Day convolution at all is not yet conclusive, since we may still try to Day convolve with respect to $(\mathbb{N},+)$. In either case, the non-trivial braiding of spheres implies that at neither stage is there enough information retained to reproduce the correct symmetric braiding.

    Karol, that’s a good point. In the spirit of MMSS00 we’d be looking at two consecutive restrictions, along the two maps

    $$\mathbb{N} \longrightarrow StdSpheres \longrightarrow \infty Grpd^{\ast/}_{fin}$$

    (where by $StdSpheres$ I mean $\mathbb{N}$ equipped with its enrichment by $StdSpheres(m,n) = S^{max(n-m,0)}$). While on $StdSpheres$ the structure maps of the sequential spectra are encoded by the enrichment, it would be understood that over $\mathbb{N}$ the structure maps are added in by hand using the monoidal structure.

    Since we need not necessarily ask that the smash product of spectra be strictly preserved along the restriction maps, only that we may somehow reconstruct it up to equivalence, maybe the failure of $StdSpheres$ to support a Day convolution at all is not yet conclusive, since we may still try to Day convolve with respect to $(\mathbb{N},+)$.

    In either case, the non-trivial braiding of spheres implies that at neither stage is there enough information retained to reproduce the correct symmetric braiding.

14. • CommentRowNumber14.
    • CommentAuthorKarol Szumiło
    • CommentTimeMar 7th 2016
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    Author: Karol Szumiło
    Format: MarkdownItexI think I misunderstood what you meant by $(\mathbb{N}, +)$. Is it just the discrete category of natural numbers with $+$ as the monoidal product?

    I think I misunderstood what you meant by $(\mathbb{N}, +)$. Is it just the discrete category of natural numbers with $+$ as the monoidal product?

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeMar 7th 2016
    • (edited Mar 7th 2016)
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    Author: Urs
    Format: MarkdownItexYes, I am writing here $\mathbb{N}$ for just the discrete category, following MMSS00, and $StdSpheres$ for the enriched version which we discussed above. Sorry if that caused confusion.

    Yes, I am writing here $\mathbb{N}$ for just the discrete category, following MMSS00, and $StdSpheres$ for the enriched version which we discussed above. Sorry if that caused confusion.

16. • CommentRowNumber16.
    • CommentAuthorKarol Szumiło
    • CommentTimeMar 8th 2016
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    Author: Karol Szumiło
    Format: MarkdownItexOK, indeed I misunderstood the notation and I wrote my original post thinking that $\mathbb{N}$ stood for what you call $StdSpheres$. I see now what you were saying. Just to summarize my point briefly: sequential spectra do not have a canonical smash product, not even a non-symmetric one, but there are of course various not quite functorial constructions.

    OK, indeed I misunderstood the notation and I wrote my original post thinking that $\mathbb{N}$ stood for what you call $StdSpheres$. I see now what you were saying.

    Just to summarize my point briefly: sequential spectra do not have a canonical smash product, not even a non-symmetric one, but there are of course various not quite functorial constructions.

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeMar 8th 2016
    • (edited Mar 8th 2016)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI have expanded and edited the wording of the remark I had written, [here](https://ncatlab.org/nlab/show/smash+product+of+spectra#WhySequentialSpectraHaveNoSymmetricSmashProduct). If it appears that I keep not expressing myself well, please feel invited to try your hands on it.

    I have expanded and edited the wording of the remark I had written, here. If it appears that I keep not expressing myself well, please feel invited to try your hands on it.

18. • CommentRowNumber18.
    • CommentAuthorMike Shulman
    • CommentTimeMar 8th 2016
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    Author: Mike Shulman
    Format: MarkdownItexThanks very much! I edited a bit more to clarify what was meant by "equivalently".

    Thanks very much! I edited a bit more to clarify what was meant by “equivalently”.

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeMar 8th 2016
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    Author: Urs
    Format: MarkdownItexOkay, thanks Mike, thanks Karol.

    Okay, thanks Mike, thanks Karol.