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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 30th 2019
• (edited Apr 30th 2019)

This entry to record the classical theorem 6.1 from

based on

I have a question:

The definition of the wedge product which they use (highlighted here) does not include the usual normalization factor of 1/2.

Question: If with this normalization we look at the quasi-isomorphism between the Sullivan model and the algebra of polynomial differential forms, do we pick up a relative factor of 2 (or 1/2) for each wedge factor?

I’ll ask the same question specialized to an explicit example in the next comment…

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeApr 30th 2019
• (edited Apr 30th 2019)

Here is the question

[edit: never mind, figured it out]

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