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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJul 11th 2014
    • (edited Jul 11th 2014)

    at normed division algebra it used to say that “A normed field is either \mathbb{R} or \mathbb{C}. ” I have changed that to “a normed field over \mathbb{R} is…” and changed normed field from being a redirect to “normed division algebra” to instead being an entry on its own.

    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeJul 11th 2014

    I added more examples to normed field.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJul 11th 2014
    • (edited Jul 11th 2014)

    Thanks for the edits!

    Regarding the edits on the links:

    I am thinking that “archimedean” and “non-archimedean” should eventually point to something other than archimedean field, an entry both more focused – in that it immediately mentions the ultrametric inequality – and more general – in that it includes not just normed fields but also normed vector spaces more generally.

    Wikipediae has “Archimedean property”. Hm.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeDec 14th 2016
    • (edited Dec 15th 2016)

    at normed division algebra in the Classification section it suddenly says “…exactly four finite-dimensional normed division algebras…”. This makes it sound as if there might be further normed division algebras that evade the classification by being infinite-dimensional. Should the “finite dimensional” qualifier not just be removed in this sentence?

    • CommentRowNumber5.
    • CommentAuthorDavidRoberts
    • CommentTimeDec 14th 2016

    John Baez has some stuff about this that we discussed once. I can track it down. Possibly it’s on MathOverflow.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 14th 2016

    Re #4: I think that’s right, but to rule out infinite-dimensional examples might require more functional analysis than is currently discussed in the relevant nLab articles. (Not much, but some.)

    • CommentRowNumber7.
    • CommentAuthorDavidRoberts
    • CommentTimeDec 14th 2016
    • (edited Dec 14th 2016)

    OK, you were right. See this MO answer and JB’s comment. I’ve edited the entry to include.

    EDIT: one has to have characteristic 2, and the algebra needs to be a purely inseparable extension of the base field, in order to get an infinite-dimensional example in the unital case – or else one needs a non-unital algebra (examples are known over the reals in this case).

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeDec 15th 2016

    Thanks! I have added one more sentence to your addition, just for emphasis of the punchline.

    • CommentRowNumber9.
    • CommentAuthorDavid_Corfield
    • CommentTimeDec 15th 2016
    • (edited Dec 15th 2016)

    Over the complex numbers, the only normed division algebra is the algebra of complex numbers themselves.

    Does this have implications if and when you try to take the superpoint story (superspacetime and branes) in intergeometric fashion over to the complex analytic world?

    What happens if you try to find maximal invariant central extensions of 0|2\mathbb{C}^{0|2}, say?

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeDec 15th 2016
    • (edited Dec 15th 2016)

    What happens if you try to find maximal invariant central extensions of 0|2\mathbb{C}^{0\vert 2}, say?

    By the same simple argument as for the real case (here) the maximal central extension of 0|2\mathbb{C}^{0\vert 2} is 3|2\mathbb{C}^{3\vert 2}, with the odd-odd bracket the one that takes two 2-component complex vectors ψ\psi, ϕ\phi to the symmetric complex matrix ψϕ t+ϕψ t\psi \phi^t + \phi \psi^t.

    This does not seem to have any good spacetime meaning. The reason why the same simple step works magic over the reals is that over the reals symmetric matrices are hermitian (trivially, but crucially), and it is hermitian 2×22 \times 2 matrices over the real normed division algebras 𝕂\mathbb{K} which are identified with Minkowski spacetimes of dimension 2+dim (𝕂)2 + dim_{\mathbb{R}}(\mathbb{K}) (here).

    • CommentRowNumber11.
    • CommentAuthorDavid_Corfield
    • CommentTimeDec 15th 2016

    This does not seem to have any good spacetime meaning.

    Which takes me back to the g+ question I posed to you about what counts as a ’valuable’ departure from a good spacetime meaning, as in F-theory. I guess there could be all kinds of ’weird’ prequantum theories which possess equivalent quantizations to more ’meaningful’ ones, as at duality in physics.

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeDec 15th 2016

    The F-theory super Lie algebra as in section 7 of arXiv:1611.06536 just so happens to be a useful correspondence space between two actual spacetimes, a “non-perturbative T-fold”. One should not read more into this than what it is: a useful book-keeping tool, for T-duality and, in this case, also for S-duality. Generally, these “doubled” T-fold spacetimes are by design not meant to be actual spacetimes, but correspondence spaces witnessing duality between actual spacetimes. They are not part of the bouquet, they are an auxiliary construction surrounding it.

    • CommentRowNumber13.
    • CommentAuthorDavid_Corfield
    • CommentTimeDec 15th 2016

    Not a task for now, but maybe you could tell us one day what’s so special about an “actual spacetime”. At spacetime there is

    A spacetime is a manifold that models space and time in physics.

    and this is said to require a smooth Lorentzian space equipped with a time orientation.

    Can we ’transcendentally’ deduce from what we experience the need for a fundamental physical theory to refer to such a spacetime?

    As I say, not something for the working day, but rather late night speculative chat.

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeDec 2nd 2018
    • (edited Dec 2nd 2018)

    There is a gap in the entry normed division algebra:

    Ever since rev 1 it claims that a normed division algebra is an algebra that is both a division algebra and a normed algebra.

    But a priori normed algebras satisfy only

    (1)|ab|C|a||b|. (1)\;\; {\vert a \cdot b\vert} \leq C {\vert a \vert} \cdot {\vert b \vert} \,.

    wheras all texts I checked define “normed division algebra” by directly imposing instead the stronger clause

    (2)|ab|=|a||b| (2)\;\;{\vert a \cdot b\vert} = {\vert a \vert} \cdot {\vert b \vert}

    I gather the idea is that with the division algebra property (1) implies (2).

    [edit: oh, of course rev 1 claims just that ]

    But this seems a bit fiddly. Or maybe I am missing something.

    • CommentRowNumber15.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 2nd 2018

    I think the necessary repairs would probably be at hand provided that we are dealing with the finite-dimensional case only. But it would take some time, which I don’t have at the moment, to go through it all. You’re right that there is some messiness here that needs cleaning up (and some of it might be non-trivial).

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeDec 2nd 2018

    Thanks, Todd! No rush. Maybe we could for the moment just add a line to the entry that confirms that the intended implication exists.

    Myself, have to rush off now.

    • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeDec 3rd 2018

    expanded the first lines as follows – this still needs attention:


    A normed division algebra is a not-necessarily associative algebra, over some ground field, that is

    1. a division algebra (i.e.(ab=0)(a=0orb=0))\big( \text{i.e.}\, (a \cdot b = 0) \Rightarrow (a = 0 \,\text{or}\, b = 0) \big)

    2. a multiplicatively normed algebra (i.e.abCab)\big( \text{i.e.}\, {\Vert a \cdot b\Vert} \leq C \cdot {\Vert a\Vert} \cdot {\Vert b\Vert} \big).

    It should be the case (at least maybe for finite-dimensional algebras) that the division property (1) implies that the norm property (2) holds in the stronger form

    |ab|=|a||b| {\vert a \cdot b\vert} \;=\; {\vert a \vert} \cdot {\vert b \vert}

    and this is how most (or all) authors actually define normed division algebras, and that’s what we assume to be meant now.

    diff, v27, current

    • CommentRowNumber18.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 4th 2018

    I mean to get back to this, but here are a few notes.

    I think it’s not that property 2. gets strengthened in this way assuming property 1. All that 2. is really trying to say is that multiplication is (uniformly) continuous with respect to the topology induced by the norm. It’s more that we always have an option to rescale the norm by putting |a|=Ca{|a|} = C {\|a\|}, so that at least |ab||a||b|{|a \cdot b|} \leq {|a|} \cdot {|b|} (called submultiplicativity).

    In the associative unital case, we can further enforce |1|=1{|\mathbf{1}|} = 1, by pulling back the operator norm on the space of continuous linear maps hom(A,A)\hom(A, A) along the map Ahom(A,A)A \to \hom(A, A) that you get by currying the multiplication map on AA. If AA has zero divisors, then the pulled-back pseudonorm does not satisfy separability (|a|=0a=0{|a|} = 0 \Rightarrow a = 0), but with property 1. you do get a norm.

    In that case, I believe the Frobenius theorem that characterizes finite-dimensional Banach algebras over \mathbb{R} as \mathbb{R}, \mathbb{C}, \mathbb{H} does give you strict multiplicativity over submultiplicativity, i.e., the equality that everyone else assumes. I’m afraid I don’t see an easy argument for getting that otherwise, and I don’t know what to make of the situation if associativity is not assumed.

    Another thing is that the article seems to goof up later on, inadvertently assuming associativity when it begins talking about Banach algebras. I assume that was unintended.

    It may be that all this fuss about starting with weaker assumptions than strict multiplicativity is just not worth it in the end, but I’m curious whether I’ve overlooked something. Maybe the author (Toby?) sees how to derive that after all. For that reason, I haven’t implemented any changes.

    • CommentRowNumber19.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 4th 2018

    Obviously in the 4th paragraph I forgot to use the word “division”.

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