Does Closure Under Multiplication in One Subspace Imply the Same for Another?

[mnb96]

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]

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.

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 $1$.

 

[mnb96]

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]

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.