Bilinear map

A bilinear map is a map $B:V\times W\to U$ between vector spaces that is linear in each argument when the other is fixed. The nonlinear temperature advection can be written as the diagonal of a bilinear map, $\mathcal{N}(\varphi ,\chi )=-\mathbf{u}[\varphi ]\cdot \nabla \chi$, when velocity depends linearly on temperature. Unlike a bilinear form, its output need not be a scalar.

A map $B:V\times W\to Z$ between vector spaces is bilinear when it is linear in either argument with the other held fixed. A scalar-valued bilinear form is the case $V=W$ and $Z$ is the scalar field. Checking bilinearity is one of the algebraic steps in verifying a real inner product; being a symmetric bilinear form and having positive squared norm impose further conditions.

Let $S:W\times W\to W$ be a symmetric bilinear map on a finite-dimensional real vector space. If $S(v,v)$ is parallel to $v$ for every $v$, then it has the displayed form for one covector $V$. Write $S(e_i,e_i)={\alpha }_i{e}_i$ in a basis. Applying the condition to $e_i+e_j$ and $e_i-e_j$ excludes every component outside their span and gives $S(e_i,e_j)=({\alpha }_j{e}_i+{\alpha }_i{e}_j)/2$. Set $V(e_i)={\alpha }_i/2$ and extend by bilinearity. Dimension one follows by the same definition. The trace identity $S^a{}_{ac}=(n+1)V_c$ proves uniqueness without depending on the chosen basis.

A continuous bilinear map obeys $\Vert f(h,k)\Vert \le C\Vert h\Vert \Vert k\Vert$. Expanding $f(a+h,b+k)$ leaves the displayed linear term and remainder $f(h,k)$. For product norm $\rho =(\Vert h{\Vert }^2+\Vert k{\Vert }^2{)}^{1/2}$, the remainder has norm at most $C{\rho }^2/2$, hence is $o(\rho )$. This proves the Fréchet derivative formula.

Let $U,V$ be nonzero finite-dimensional complex vector spaces, and let $\varphi :U\otimes V\to W$ have injective restrictions when either nonzero factor is fixed. It induces a map from ${\mathbb{CP}}^{a-1}\times {\mathbb{CP}}^{b-1}$ to ${\mathbb{CP}}^{r-1}$, where $a=\dim U$, $b=\dim V$, and $r=\mathrm{rank}\varphi$. Pullback of the degree-two generator is $x+y$, since both factor restrictions are projective linear embeddings. Its $(a+b-2)$th power is $\left(\genfrac{}{}{0ex}{}{a+b-2}{a-1}\right)x^{a-1}{y}^{b-1}\ne 0$ in integral cohomology. The target must therefore have complex dimension at least $a+b-2$. Positivity of both factor dimensions is necessary; a zero factor can make the claimed inequality false.

For nonzero finite-dimensional complex vector spaces $U,V$, suppose a bilinear map never vanishes on a pair of nonzero vectors. Equivalently, each map obtained by fixing one nonzero argument is injective. If its associated map on $U\otimes V$ has image dimension $r$, projectivization gives a map to ${\mathbb{CP}}^{r-1}$ whose positive degree-two class pulls back to $x+y$. The nonzero top power $(x+y{)}^{\dim U+\dim V-2}$ in the product cohomology ring of complex projective space proves the bound. Multiplication of polynomials of bounded degree attains equality. Nonzero hypotheses matter: with a zero factor, slice conditions can be vacuous.