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Given $ a= (x_1,x_2), b= (y_1,y_2) \in \mathbb{C}^2 $ and $f(a,b) = x_1\bar y_2 - \bar x_2 y_1$

Now my question: Is $f$ a bilnear form on $\mathbb{C}^2$ ?

My attempt: i know that for a bilinear form, we have $f(ka_1 +b_1 , c_1) = k f(a_1,c_1) + f(b_1, c_1)$.

But here I don't know how to check whether $f$ is bilnear form or not ?

Any hints/solution will be appreciated

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  • $\begingroup$ A bilinear form should also have $f(a_1,kb_1+c_1)=kf(a_1,b_1)+f(a_1,c_1)$. I can't wait to see those $\bullet_1$ subscripts come back to bite. $\endgroup$ – Gae. S. Nov 9 '19 at 9:07
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    $\begingroup$ @Gae.S. Isn't a complex "bilinear" form conjigate linear in one of it's arguments, i.e. sesquilinear? $\endgroup$ – Botond Nov 9 '19 at 9:19
  • $\begingroup$ @Botond You're answering yourself: "sesquilinear", as opposed to "bilinear". $\endgroup$ – Gae. S. Nov 9 '19 at 9:20
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Let $a=b=(1,1)$ and $\lambda \in \mathbb{C} - \mathbb{R}$, then

$f(a,\lambda b)= \bar{\lambda} - \lambda \neq 0=f(a,b)$.

Thus if the scalar fields of $\mathbb{C}^2$ is $\mathbb{C}$, then the answer is NO.

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Take $\alpha$ complex. We check linearity in the first argument.

In general $f(\alpha a,b)=\alpha x_1 \overline{y_2}-\overline{\alpha}\overline{x_2}y_1$.

If we take $x_1=x_2=y_1=y_2=1$ than $f(a,b)=0$ but $f(\alpha a,b)=2i Im(\alpha)$. This shows that the operation is not bilinear, not being even linear in the first argument.

UPDATE: This shows that the function cannot be a scalar product over $\mathbf{C}^2_{\mathbf C}$. That is if we see $\mathbf{C^2}$ as a vector space over $\mathbf{C}$. If we see it as a vector space over $\mathbf{R}$, that is $\mathbf{C}^2_{\mathbf R}$, than we just allow multiplication by real numbers the operation $f$ ( which is now a map $f: \mathbf{C}^2_{\mathbf R} \times \mathbf{C}^2_{\mathbf R} \rightarrow \mathbf{C}_{\mathbf R}$ ) in that case should be bilinear (in this sense: https://en.wikipedia.org/wiki/Multilinear_map ). To see this quickly I suggest to use that the scalar product is equal to the determinant of a particular 2x2 matrix and use the multilinearity property of the determinant with respect to linear operations on the columns ...

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