# Subalgebras and direct sums

• A

## Main Question or Discussion Point

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).

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fresh_42
Mentor
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 $1$.

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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
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.