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A Subalgebras and direct sums

  1. May 18, 2018 #1
    Hi,

    consider a (finite dimensional) vector space ##V=U\oplus W##, where the subspaces ##U## and ##V## are not necessarily orthogonal, equipped with a bilinear product ##*:V\times V \rightarrow V##.

    The subspace ##U## is closed under multiplication ##*##, thus ##U## is a subalgebra of ##V##.

    Does this imply that also ##W## is a subalgebra of ##V##?

    (Note, I can already prove the special case that if U and W are orthogonal, then both U and W are indeed subalgebras).
     
  2. jcsd
  3. May 18, 2018 #2

    fresh_42

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    Just as a side note: orthogonal doesn't make sense, as long as you don't specify the quadratic form and the field. Vector spaces in general don't automatically allow inner products.

    The answer to your question is no. Example:
    ##h=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\; , \;x=\begin{bmatrix}0&1\\0&0\end{bmatrix}\; , \;y=\begin{bmatrix}0&0\\1&0\end{bmatrix}## with ##V=\operatorname{span}_\mathbb{F}\{\,h,x,y\,\}\; , \;U=\mathbb{F}\cdot h\; , \; W=\operatorname{span}_\mathbb{F}\{\,x,y\,\}##. With the multiplication ##v*w= v\cdot w - w \cdot v## we have ##h*h=0\in U## and ##x*y=h \notin W##.

    Edit: Typo corrected. ##h_{21}=0## not ##1##.
     
    Last edited: May 22, 2018
  4. May 22, 2018 #3
    Hi fresh_42,

    you gave a very interesting counterexample of my statement that is actually too inspiring to close the discussion here :)

    In fact, let's define the "product of two subspaces" as ##UV=\left \{uv\;|\; u\in U, \, v\in V \right \}##, and notice that in your construction ##H^2=0##. In other words, ##H## (as a set) acted as a nilpotent w.r.t. the product.

    I am wondering if it is possible to find a similar counterexample, in which ##H^2=H##, i.e. the closure of ##H## w.r.t. the product is ##H## itself.
     
  5. May 22, 2018 #4

    fresh_42

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    It is the Lie algebra ##\mathfrak{sl}(2)## with ##.*. =[.,.]## as Lie multiplication.
    What does "as a set" mean? ##H=h\cdot \mathbb{F}\,##? That's the heritage of the Lie algebra structure, where ##[X,X]=X*X=0## holds for any element.
    We don't have any restrictions for the multiplication. So we can simply define a multiplication by ##A^2=A## and leave all other as they are: ##H*X=2X, H*Y=-2Y,X*Y=H##. I don't see any obvious reasons, why this shouldn't work. However, to find a realization by matrices or similar could take a moment, at least if we don't want to use the tensor algebra and its universal property. I would look among genetic algebras for an example.
     
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