# Subalgebras and direct sums

• A
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).

fresh_42
Mentor
2021 Award
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 that if U and W are orthogonal, then both U and W are indeed subalgebras, but I am interested in the general case).
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.

##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:
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.

fresh_42
Mentor
2021 Award
Hi fresh_42,

you gave a very interesting counterexample of my statement that is actually too inspiring to close the discussion here :)
It is the Lie algebra ##\mathfrak{sl}(2)## with ##.*. =[.,.]## as Lie multiplication.
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.
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.
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.
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.