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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeDec 8th 2011
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   Author: Andrew Stacey
   Format: MarkdownItexThought I'd write up some old notes at [[symmetric product of circles]] (linked from [[unitary group]], explanation to come on [[symmetric product of circles]]). Not finished yet, but have to leave it for now. (I was incensed to discover that to look at the source article for the material for this to check that I'm remembering it right - I last looked at it about 10 years ago - I have to pay 30 UKP. The article is 3 pages long. That's 10UKP per page! So I'm going from vague memories and "working it out afresh".)

   Thought I’d write up some old notes at symmetric product of circles (linked from unitary group, explanation to come on symmetric product of circles). Not finished yet, but have to leave it for now.

   (I was incensed to discover that to look at the source article for the material for this to check that I’m remembering it right - I last looked at it about 10 years ago - I have to pay 30 UKP. The article is 3 pages long. That’s 10UKP per page! So I’m going from vague memories and “working it out afresh”.)

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 8th 2011
   • (edited Dec 8th 2011)
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   Author: Todd_Trimble
   Format: MarkdownItexAndrew, I'm confused by your definition 1. I expect what you mean is to take the joint coequalizer of the maps $X^n \to X^n$ induced by permuting coordinates, and this can be phrased equivalently as the coequalizer of the _pair_ of maps $$\coprod_{\sigma \in \Sigma_n} X^n \cong S_n \times X^n \stackrel{\overset{\pi}{\to}}{\underset{\alpha}{\to}} X^n$$ where $\pi$ is projection onto the second coordinate, and $\alpha$ is the canonical $S_n$-action. Is that what you meant?

   Andrew, I’m confused by your definition 1. I expect what you mean is to take the joint coequalizer of the maps $X^n \to X^n$ induced by permuting coordinates, and this can be phrased equivalently as the coequalizer of the pair of maps

   $$\coprod_{\sigma \in \Sigma_n} X^n \cong S_n \times X^n \stackrel{\overset{\pi}{\to}}{\underset{\alpha}{\to}} X^n$$

   where $\pi$ is projection onto the second coordinate, and $\alpha$ is the canonical $S_n$-action. Is that what you meant?

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeDec 8th 2011
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   Author: Andrew Stacey
   Format: MarkdownItexTodd, probably! I was trying to be "fancy" and put a categorical definition (this is the nlab, after all). I was a bit surprised that we didn't have a page [[symmetric product]] already as it seems a fairly obvious notion.

   Todd, probably! I was trying to be “fancy” and put a categorical definition (this is the nlab, after all). I was a bit surprised that we didn’t have a page symmetric product already as it seems a fairly obvious notion.

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeDec 9th 2011
   • (edited Dec 9th 2011)
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   Author: TobyBartels
   Format: MarkdownItexShouldn't the term be [[symmetric power]]? We have an article on that (with a definition that respects multiplicities), albeit restricted to a different context. (Probably that article should be left as it is, at [[symmetric algebra]], with a new page [[symmetric power]] for the concept in an arbitrary symmetric monoidal category.)

   Shouldn’t the term be symmetric power? We have an article on that (with a definition that respects multiplicities), albeit restricted to a different context. (Probably that article should be left as it is, at symmetric algebra, with a new page symmetric power for the concept in an arbitrary symmetric monoidal category.)

5. • CommentRowNumber5.
   • CommentAuthorAndrew Stacey
   • CommentTimeDec 9th 2011
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   Author: Andrew Stacey
   Format: MarkdownItexI think that in topology it's more usual to talk of the symmetric product. Nonetheless, as we have the page [[symmetric power]] then I took out the general definition from this page and linked to that page. Also completed the proof of the description of $SP^n(S^1)$ and added that of $SP^n(\mathbb{R})$. Still to do is the connection to unitary matrices.

   I think that in topology it’s more usual to talk of the symmetric product. Nonetheless, as we have the page symmetric power then I took out the general definition from this page and linked to that page.

   Also completed the proof of the description of $SP^n(S^1)$ and added that of $SP^n(\mathbb{R})$. Still to do is the connection to unitary matrices.

6. • CommentRowNumber6.
   • CommentAuthorjim_stasheff
   • CommentTimeDec 9th 2011
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   Author: jim_stasheff
   Format: TextYes, in topology it's called the symmetric product. Recent flurry of activity over at algtop mailing list in re: cells for SP^infty(S^n)

   Yes, in topology it's called the symmetric product. Recent flurry of activity over at algtop mailing list
   in re: cells for SP^infty(S^n)