nate
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Post by nate on Oct 2, 2020 6:55:56 GMT
Thanks for the quick reply. You know, there is another reason why I really like the form $\partial_x + i\partial_y$, but it might be too far afield to be relevant here. Recently I was reading through some of Spivak's *Introduction to Differential Geometry*. In the discussion of the tangent bundle (Ch. 3, page 78 in my book), he shows, quite elegantly, that the space of all linear operators which are "derivations at $p$" is spanned by the familiar partial derivatives. Somehow, this rings a bell with the form of the equation with $\partial_x + i \partial_y$.
It also highlights a nice fact (I'm not sure where it is in the book...), that exposes a way of looking at complex multiplication. If $z = x + iy$, $\mathbb{z} = (x,y)$, $w = u + iv$, and $\mathbb{w} = (u,v)$, then
$$z \cdot w = \langle\mathbb{z},\mathbb{\bar w}\rangle - i\mathbb{z}\times\mathbb{\bar w}$$
While not as useful in general as considering complex multiplication to be a dilative rotation, it does at least bring complex multiplication into terms of familiar vector operations, and makes plausible (at least from a purely formal perspective) the formula
$$\partial_x + i\partial_y = \langle\nabla, \bar{\cdot}\rangle - i\nabla \times \,\bar{\cdot}$$
It may very well be that this is all balderdash, but I can't help but feeling there is some useful conceptual tool or somewhat interesting connection buried here somewhere...
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Post by Admin on Oct 5, 2020 8:57:14 GMT
Hi nate
Interesting, but what do the brackets $\langle\rangle$ mean here?
Thanks
Vasco
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nate
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Post by nate on Oct 5, 2020 9:08:46 GMT
Here the $\langle,\rangle$ is meant to mean "dot product."
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Post by Admin on Oct 5, 2020 9:37:19 GMT
Here the $\langle,\rangle$ is meant to mean "dot product." Hi nate So $\langle\boldsymbol{a},\boldsymbol{b}\rangle\equiv\boldsymbol{a\cdot b}$. Is that right? Vasco
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nate
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Post by nate on Oct 5, 2020 9:40:09 GMT
Yes sir. I didn't wish to use the "dot" notation to distinguish it from complex multiplication.
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Post by mondo on Apr 14, 2022 7:06:41 GMT
In the last equation of page 486 after the second equal sign, why $x$ was replaced with $\frac{1}{2}(z+\overline{z})$?
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Post by Admin on Apr 14, 2022 9:12:40 GMT
In the last equation of page 486 after the second equal sign, why $x$ was replaced with $\frac{1}{2}(z+\overline{z})$? Mondo Do you see that $x$ is identically equal to $\frac{1}{2}(z+\overline{z})$? Note that $z+\overline{z}=x+iy+x-iy=2x$ Then this substitution enables us to write (17) in terms of $z$ and $\overline{z}$ Vasco
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Post by mondo on Apr 15, 2022 6:45:02 GMT
Ahh right, that is even visible from a vector addition of two vectors reflected in the real axis. Thank you.
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