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    • CommentRowNumber1.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    I wrote partial differentiation. Also some remarks at chain rule.

    • CommentRowNumber2.
    • CommentAuthorzskoda
    • CommentTimeJun 7th 2013

    You use link differentiable space which so far does not exist in nnLab. On the other hand, this is strange as there are so many entries on alike topics with many redirects. I am convinced that the entry with the content which you meant is somewhere, just should be properly referred to, don’t you think so ? (I am not sure what formalism you precisely meant)

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJun 7th 2013
    • (edited Jun 7th 2013)

    Okay, so just in order to satisfy links in partial differentiation, I have now created brief entries

    and made product projection redirect to projection.

    Also added links from differentiation and differential equation.

    • CommentRowNumber4.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 7th 2013

    Do we really need the additive structure?

    Isn’t the statement “Then under good conditions, we have …” really saying “Then under good conditions the differentiation functor is product preserving.” That sum could just as well be the product.

    I’m also not keen on the following paragraph:

    More precisely, we have a category of differentiable spaces and differentiable maps, on which is an endofunctor that takes each space UU to a notion of tangent bundle TUT{U}, which is a vector bundle over UU, and takes a map f:UYf\colon U \to Y to df:TUTYd{f}\colon T{U} \to T{Y}. Then dx i:TUTX id{x_i}\colon T{U} \to T{X_i}, if p:T x i(p)X iT f(p)Y\partial_i{f}_p\colon T_{x_i(p)}{X_i} \to T_{f(p)}{Y} is a linear operator between stalks (for pp a point in UU), and the sum takes place in the vector space T f(p)YT_{f(p)}{Y}.

    “we have a category of differentiable spaces” makes it sound as though there is just one such, or there is one obvious one before which all others must bow.

    Even were that the case, the tangent bundle need not be either a bundle or have fibres vector spaces unless our category is just something like smooth manifolds.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJun 7th 2013
    • (edited Jun 7th 2013)

    As a quick hack, I have changed it to

    More precisely, we have a category of differentiable spaces (for instance microlinear spaces)

    but it would be good if somebody found the time and energy for a more comprehensive paragraph.

    But if so, please think about if such general discussion is not better had at differentiation itself.

    • CommentRowNumber6.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 7th 2013

    I’ve removed the remark about kids’ tests from chain rule, it jarred for me with the general tone of the nLab.

    I’ve added in a segment about partial differentiation in a cartesian closed category of smooth spaces, but I have to go soon so can’t develop it further for now. What I want to say is that if the tangent functor is product preserving (see Proposition 4.4 of my Yet More Smooth Mapping Spaces … paper) then you can reconstruct the full derivative from the partials. But I’m not sure I quite understand the correct statement of that right now.

    • CommentRowNumber7.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    “we have a category of differentiable spaces” makes it sound as though there is just one such

    No, I meant nothing of the sort! That's why I used ‘a’ instead of ‘the’. The context should be fairly generic.

    • CommentRowNumber8.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    Also added links from differentiation and differential equation.

    I didn't see the latter, so I put one in, also justifying the redirect from partial differential equation.

    • CommentRowNumber9.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    Andrew, I like what you've done!

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 7th 2013

    The little word ’a’ can definitely confuse on occasion! Here, the meaning seems to be synonymous with “we have categories of differentiable spaces”.

    • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013
    • (edited Jun 7th 2013)

    Ah, I see! It's really a question of the scope of ‘have’. The current ‘choose’ is an improvement.