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Let $V$ be a (finite dimensional) vector space over $K$, and $B: V \times V \to K$ a bilinear form. Consider a decomposition $V=V_1 \oplus V_2$, into vector subspaces $V_1,V_2$ of $V$.

In general, if we have bilinear forms $B_1, B_2$ on vector spaces $V_1,V_2$, then we have a bilinear form $\hat{B}: V_1 \oplus V_2 \times V_1 \oplus V_2 \to K$ defined by $\hat{B}((x_1,y_1),(x_2,y_2))=B_1(x_1,x_2)+B_2(y_1,y_2)$

Now clearly $B$ restricts to a bilinear form on each of $V_1,V_2$. Then it's natural to identify $\hat{B}$ (induced by $B|_{V_1}$ and $B|_{V_2}$) with $B$ on $V$ (right??)

But now let $x \in V_1, y\in V_2$ then $B(x,y)=\hat{B}((x,0),(0,y))=B(x,0)+B(0,y)=0$, which does not make sense as $B$ is just an arbitrary bilinear form on $V$.

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    $\begingroup$ The bilinear form $B$ together with the direct sum decomposition $V_1 \oplus V_2$ determines (via restriction) four bilinear maps $B_{ij} := B\vert_{V_i \times V_j} : V_i \times V_j \to K$. Of course, only the maps $B_{ii}$ are bilinear forms (on $V_i$, respectively). Likewise, maps $B_{ij}$ together determined a bilinear form $B$. The presentation in the 'question' here seems to ignore the maps $B_{12}$ and $B_{21}$, which in general need not be zero. $\endgroup$ – Travis Mar 6 '17 at 15:39
  • $\begingroup$ @Travis I see, thanks! So I was wrong to think of $B$ as simply $B|_{V_1} \oplus B|_{V_2}$, which was quite silly of me for the reasons you pointed out... $\endgroup$ – AlexTaylor Mar 6 '17 at 15:57
  • $\begingroup$ You're welcome. For an extreme example of this, consider the skew bilinear form $\omega((x_1, y_1), (x_2, y_2)) := x_1 y_2 - x_2 y_1$ on $\Bbb R^2 \leftrightarrow \Bbb R \oplus \Bbb R$, which we can identify with the standard volume form. Then, the restriction of $\omega$ to each factor $\Bbb R$ is zero, but $\omega$ is itself not zero. $\endgroup$ – Travis Mar 6 '17 at 16:03

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