Coproduct in the category of vector spaces with bilinear forms

Question

I'm trying to work out the coproduct in the category of (say real) vector spaces equipped with bilinear forms, where the morphisms $(V,b) \to (V',b')$ are the linear maps $T : V \to V'$ such that $T^* b' = b$, that is, $b'(Tv,Tw) = b(v,w)$. It seems like it should be $(V \oplus V', b \oplus b')$ where we define $(b \oplus b')((v,v'),(w,w')) = b(v,w)+b'(v',w')$. However, if we take $(V,b)$ and $(V',b')$ to both be $\mathbb{R}$ with the usual bilinear form $b(x,y) = xy$, and apply the universal property to the pair of morphisms $\text{id} : \mathbb{R} \to \mathbb{R}$ and $\text{id} : \mathbb{R} \to \mathbb{R}$, then this should give rise to a unique morphism $h : \mathbb{R}^2 \to \mathbb{R}$ satisfying $(x+y)(x'+y') = xx' + yy'$. But this is not true since there are cross terms in $(x+y)(x'+y')$. What am I doing wrong?

Answer 1

Notice that $V$ and $V'$ are orthogonal with respect to $b \oplus b'$, i.e. we have $(b \oplus b')(v \oplus v')=0$ and $(b \oplus b')(v' \oplus v)=0$ for all $v \in V$, $v' \in V'$. In fact, $(V \oplus V',b \oplus b')$ represents the subfunctor of $\hom((V,b),-) \times \hom((V',b'),-)$ which consists of those pairs of morphisms $f : (V,b) \to (W,c)$, $g : (V',b') \to (W,c)$ such that $c(f(v),g(w))=c(g(w),f(v))=0$ for all $v \in V$, $w \in W$. Accordingly, you may view $(V \oplus V',b \oplus b')$ as the orthogonal direct sum of $(V,b)$ of $(V',b')$. Some authors omit the word "orthogonal" here (which might be confusing). As you have observed, $(V \oplus V',b \oplus b')$ is not a coproduct.

Notice that in this category, let's call it $\mathsf{Bilin}$, the object $(0,0)$ is initial, but it is not terminal. In fact, there is a morphism $(V,b) \to (0,0)$, and then it is unique, if and only if $b=0$. It follows that $\mathsf{Bilin}$ (unlike $\mathsf{Vect}$) has no linear structure. In fact, it is more or less a "geometric category".

The category $\mathsf{Bilin}$ has no coproducts. Assume for example that $(\mathbb{R},0) \oplus (\mathbb{R},0)$ exists, say $(U,b)$. Then we have two elements $v,w \in U$ with $b(v,v)=0$, $b(w,w)=0$, such that for every other $(W,c)$ containing two such elements $\overline{v},\overline{w}$ there is a unique linear map $f : U \to W$ such that $b(x,y)=c(f(x),f(y))$ and $f(v)=\overline{v}$, $f(w)=\overline{w}$. In particular, $b(v,w)=c(\overline{v},\overline{w})$. But there is no reason why the right hand side should be constant. It could be zero or non-zero, depending on the choices of $(W,c)$ and $\overline{v},\overline{w}$. In fact, if we consider $(0,0)$ with $\overline{v}=\overline{w}=0$, it follows $b(v,w)=0$. But if we consider $(\mathbb{R}^2,b)$ determined by $b(e_1,e_1)=0$, $b(e_2,e_2)=0$, $b(e_1,e_2)=b(e_2,e_1)=1$, and $\overline{v}=e_1$, $\overline{w}=e_2$, then it follows $b(v,w)=1$. This is a contradiction.

More generally, if $(V,b) \oplus (V',b')$ exists, it is likely that $V=0$ or $V'=0$. I think that this can be proven with the same method as above. Conversely, if $V=0$, then $(V,b)$ is the initial object and the coproduct exists of course.

A better behaved category is suggested by Zhen Lin in the comments.