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Cauchy-Riemann and Linear Independence

  1. Mar 30, 2006 #1
    Here's an interesting way to look at CR I feel is often overlooked:

    Let:

    [tex]z = x + i y[/tex]
    [tex]z^{\ast} = x - i y[/tex]

    One common form for the CR condition is to say that if some function [itex]f[/itex] is analytic then it does not depend on [itex]z^{\ast}\;[/itex]. That is,

    [tex]\frac{\partial f}{\partial z^{\ast}} = 0[/tex]

    But is there anything special about [itex]z^{\ast}\;[/itex]? It turns out there isn't. In fact, any complex variable [itex]w[/itex] that is linearly independent with [itex]z[/itex] will do. To illustrate this, let

    [tex]w = a x - i b y[/tex]

    Then,

    [tex]b z + w = (a+b)x, \, a z - w = i(a+b)y[/tex]

    So,

    [tex]x = \frac{b z + w}{a + b}, \, y = \frac{a z - w}{i(a + b)}[/tex]
    [tex]\frac{\partial x}{\partial w} = \frac{1}{a+b}, \, \frac{\partial y}{\partial w} = -\frac{1}{i(a+b)}[/tex]

    We then have

    [tex]\frac{\partial}{\partial w} = \frac{\partial x}{\partial w} \frac{\partial}{\partial x} + \frac{\partial y}{\partial w} \frac{\partial}{\partial y} = \frac{1}{a+b}[\frac{\partial}{\partial x} - \frac{1}{i} \frac{\partial}{\partial y}][/tex]

    We then confirm that effectively,

    [tex]\frac{\partial z}{\partial w} = \frac{1}{a+b}[\frac{\partial}{\partial x}(x + i y) - \frac{1}{i} \frac{\partial}{\partial y}(x + i y)] = \frac{1}{a+b}[1 - \frac{i}{i}] = 0[/tex]

    which holds as long as [itex]a+b \neq 0[/itex] which is to say that [itex]z[/itex] and [itex]w[/itex] are linearly independent. Other than this constraint, our choice of [itex]a[/itex] and [itex]b[/itex] is totally arbitrary. The CR condition is simply:

    [tex]\frac{\partial f}{\partial w} = \frac{1}{a+b}[\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}]f = 0[/tex]

    Or simply,

    [tex][\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}]f = 0[/tex]
     
    Last edited: Mar 30, 2006
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  3. Mar 30, 2006 #2
    If you have a complex variable which is ax+iby and is not simply a multiple of x+iy, then you can write it as a linear combination of z and z*. Therefore is a function is dependant on your new variable, w, then it is also dependant on z*, which (I think) leads you to the same conclusion but without having to wade through the algebra.
     
  4. Mar 30, 2006 #3

    Hurkyl

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    And here, we can even give an explicit dependence: I think that [itex]w = (a + b)z^*[/itex].

    I think, though, that it's an interesting question to forget about the complex structure, and just consider maps from R²->R², and what these derivatives look like.
     
  5. Mar 30, 2006 #4
    I think you think correctly...but...

    I guess my focus was on avoiding the use of [itex]z^\ast[/itex] entirely in the derivation of CR to illustrate what I feel is the essence of CR...linear independence, and hence an invertible map [itex](x,y) \leftrightarrow (z, w)\;[/itex].
     
  6. Mar 30, 2006 #5

    JasonRox

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    That's pretty interesting.

    I never thought about it that way.
     
  7. Mar 31, 2006 #6
    Vectors And Spinors

    A related topic I wanted to bring up was how vectors in [itex]R^2[/itex] are related to points on the complex plane and what I think the complex plane can be intuitively considered to represent geometrically. Note that this approach is just one of potentially many interpretations we could give the complex numbers - but I offer this one here since it's greatly helped me understand them better myself.

    Let [itex]\{ \mathbf{e}_1, \mathbf{e}_2 \}[/itex] be an orthonormal basis for [itex]R^2\;[/itex]. Using a Euclidean norm, we have:

    [tex]\mathbf{e}_i \cdot \mathbf{e}_j = \delta_{i j}[/tex]

    Let us go beyond the inner product and define an associative geometric product that satisfies the following rules:

    [tex]\mathbf{e}_1 \mathbf{e}_1 = \mathbf{e}_2 \mathbf{e}_2 = 1[/tex]

    [tex]\mathbf{e}_1 \mathbf{e}_2 = -\mathbf{e}_2 \mathbf{e}_1 = \mathbf{e}_{12}[/tex]

    Furthermore, let the product distribute over addition and commute with real scalars. The justification for these definitions will become clear shortly as we explore the algebraic properties of the product and what they correspond to geometrically.

    [itex]\mathbf{e}_{12}[/itex] represents an oriented unit area on the plane spanned by [itex]\{ \mathbf{e}_1, \mathbf{e}_2 \}\;[/itex]. When left-multiplied by a vector, it has the effect of rotating the vector by [itex]90^\circ[/itex] on the plane in the direction from [itex]\mathbf{e}_1[/itex] to [itex]\mathbf{e}_2[/itex] as can be seen from:

    [tex]\mathbf{e}_1 \mathbf{e}_{12} = \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 = (\mathbf{e}_1 \mathbf{e}_1) \mathbf{e}_2 = \mathbf{e}_2[/tex]
    [tex]\mathbf{e}_2 \mathbf{e}_{12} = \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 = (\mathbf{e}_2 \mathbf{e}_1) \mathbf{e}_2 = (-\mathbf{e}_1 \mathbf{e}_2) \mathbf{e}_2 = -\mathbf{e}_1 (\mathbf{e}_2 \mathbf{e}_2) = -\mathbf{e}_1[/tex]

    When right-multiplied it rotates the vector in the opposite direction:

    [tex]\mathbf{e}_{12} \mathbf{e}_1 = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_1 = (\mathbf{e}_1 \mathbf{e}_2) \mathbf{e}_1 = (-\mathbf{e}_2 \mathbf{e}_1) \mathbf{e}_1 = -\mathbf{e}_2 (\mathbf{e}_1 \mathbf{e}_1) = -\mathbf{e}_2[/tex]
    [tex]\mathbf{e}_{12} \mathbf{e}_2 = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_2 = \mathbf{e}_1 (\mathbf{e}_2 \mathbf{e}_2) = \mathbf{e}_1[/tex]

    If [itex]\mathbf{e}_1[/itex] points horizontally towards the right and [itex]\mathbf{e}_2[/itex] points vertically upwards, then [itex]\mathbf{e}_{12}[/itex] rotates vectors [itex]90^\circ[/itex] counterclockwise when left-multiplied by a vector and [itex]90^\circ[/itex] clockwise when right-multiplied by a vector.

    It should be clear then that either right- or left-multiplying a vector twice by [itex]\mathbf{e}_{12}[/itex] should rotate the vector [itex]180^\circ[/itex] and should thus negate each of its coordinates. To see this explicitly from our algebraic rules:

    [tex]\mathbf{e}_{12} \mathbf{e}_{12} = (\mathbf{e}_{1} \mathbf{e}_{2}) (\mathbf{e}_{1} \mathbf{e}_{2}) = (-\mathbf{e}_{2} \mathbf{e}_{1}) (\mathbf{e}_{1} \mathbf{e}_{2}) = -\mathbf{e}_{2} (\mathbf{e}_{1} \mathbf{e}_{1}) \mathbf{e}_{2} = -\mathbf{e}_{2} \mathbf{e}_{2} = -1[/tex]

    So [itex]\mathbf{e}_{12}[/itex] squares to -1 and commutes with real scalars, so we can identify it with a unit imaginary [itex]i\;[/itex]. So now we can think of the linear space spanned by [itex]\{1, \mathbf{e}_{12}\}[/itex] as the complex plane [itex]C[/itex] and we shall call its elements spinors to contrast with the elements of [itex]R^2\;[/itex], the linear space spanned by [itex]\{\mathbf{e}_1, \mathbf{e}_2\}[/itex] which we've been calling vectors.

    Under the geometric product:
    • a vector and a vector produce a spinor
    • a vector and a spinor produce a vector
    • a spinor and a spinor produce a spinor

    So spinors are closed under the product whereas vectors are not. Under the product, vectors map [itex]R^2 \rightarrow C[/itex] and [itex]C \rightarrow R^2\;[/itex]. By choosing a unit vector [itex]\hat{\mathbf{u}}[/itex] for the real axis, any spinor [itex]z[/itex] can be mapped to [itex]R^2[/itex] via [itex]z \rightarrow \hat{\mathbf{u}} z\;[/itex]. For our purposes here, it will be useful to let [itex]\hat{\mathbf{u}} = \mathbf{e}_1\;[/itex].

    Under the product, spinors map [itex]R^2 \rightarrow R^2[/itex] by performing a scaling and a rotation of a vector. Unit spinors are called rotors and can be thought of as the points on the complex plane lying on the unit circle. Rotors do not affect the scaling of the vectors. Spinors that scale vectors but do not rotate them are simply called, you guessed it, real scalars! These are the points on the complex plane that lie on the real axis.
     
    Last edited: Mar 31, 2006
  8. Mar 31, 2006 #7
    Vectors, Spinors, and Cauchy-Riemann

    For what follows, I shall use the following typeface convention:
    • scalars: [itex]\mathit{x}, \mathit{y}[/itex]
    • vectors: [itex]\mathbf{u}, \mathbf{v}[/itex]
    • spinors: [itex]\mathtt{z}, \mathtt{w}[/itex]
    All scalars here are assumed to be real numbers. [itex]\mathbf{e}_1[/itex] and [itex]\mathbf{e}_2[/itex] are orthonormal vectors. [itex]i = \mathbf{e}_{12}[/itex] is a unit imaginary, or unit bivector for the plane spanned by [itex]\{ \mathbf{e}_1, \mathbf{e}_2 \}\;[/itex].

    Consider some [itex]\mathbf{v} \in R^2\;[/itex],

    [tex]\mathbf{v} = \mathit{x}\, \mathbf{e}_1 + \mathit{y}\, \mathbf{e}_2\;[/tex].

    Now let's map [itex]\mathbf{v}[/itex] into [itex]C[/itex] by treating [itex]\mathbf{e}_1[/itex] as a unit vector on the real axis.

    [tex]\mathtt{z} = \mathbf{e}_1 \, \mathbf{v} = \mathbf{e}_1 (\mathit{x}\, \mathbf{e}_1 + \mathit{y}\, \mathbf{e}_2) = \mathit{x} + \mathit{y}\, \mathbf{e}_{12}[/tex]

    To map [itex]\mathtt{z}[/itex] back to [itex]\mathbf{v}\;[/itex],

    [tex]\mathbf{e}_1\; \mathtt{z} = \mathbf{e}_1 \mathbf{e}_1 \, \mathbf{v} = \mathbf{v}[/tex]

    So it is clear that spinor [itex]\mathtt{z}[/itex] will transform [itex]\mathbf{e}_1[/itex] to [itex]\mathbf{v}[/itex] when operating from the right in the expression by scaling it by some factor and then rotating it by some angle. [itex]\mathtt{z}[/itex] will transform all vectors in [itex]R^2[/itex] the same way, using the same scaling factor and rotation angle.

    Now consider the following vector operator:

    [tex]\nabla = \mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} + \mathbf{e}_2 \frac{\partial}{\partial \mathit{y}}[/tex]

    We might recognize this as the gradient operator for a 2 dimensional space with Euclidean norm.

    Something truly surprizing (or perhaps expected) occurs when we map this operator to [itex]C[/itex] using [itex]\mathbf{e}_1\;[/itex]:

    [tex]\mathbf{e}_1 \nabla = \mathbf{e}_1 (\mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} + \mathbf{e}_2 \frac{\partial}{\partial \mathit{y}}) = \frac{\partial}{\partial \mathit{x}} + \mathbf{e}_{12} \frac{\partial}{\partial \mathit{y}} = \frac{\partial}{\partial \mathit{x}} + i \frac{\partial}{\partial \mathit{y}}[/tex]

    Does this operator look familar, perchance? What if we let it act on [itex]\mathtt{z}\;[/itex]? What about [itex]\mathtt{z}^k\;[/itex]? Consider a map [itex]f: C \rightarrow C\;[/itex]. If [itex]f(\mathtt{z})[/itex] has a Taylor Series expansion in [itex]\mathtt{z}[/itex], what can we say about [itex]\nabla f\;[/itex]?
     
    Last edited: Mar 31, 2006
  9. Mar 31, 2006 #8
    So what the heck is the complex derivative anyhow?

    We've seen how we can map vectors in [itex]R^2[/itex] to spinors in [itex]C[/itex] using some simple algebraic rules defined for multiplying vectors by vectors. By selecting a unit vector lying on the "real axis" we found a unique bijective map [itex]R^2 \longleftrightarrow C\;[/itex]. Now let's look at the map [itex]f: C \rightarrow C[/itex] and apply this insight to it.

    Using [itex]\mathbf{e}_1[/itex] as our real unit vector, how would we find the corresponding map [itex]F: R^2 \rightarrow R^2\;[/itex]? Could we map a vector to a spinor argument for [itex]f[/itex] and then map the spinor result of [itex]f[/itex] back to a vector? Surely this should leave the analytic structure of [itex]f[/itex] intact, no? Like Hurkyl mentioned above, we could just look at how [itex]F[/itex] acts in [itex]R^2[/itex] without regard for the complex structure.

    So then let

    [tex]\mathbf{v} = \mathbf{e}_1 \, \mathtt{z}[/tex]

    [tex]\mathtt{z} = \mathbf{e}_1 \, \mathbf{v}[/tex]

    [tex]F(\mathbf{v}) = \mathbf{e}_1 \, f(\mathtt{z}) = \mathbf{e}_1 \, f(\mathbf{e}_1 \, \mathbf{v})\;[/tex].

    [tex]f(\mathtt{z}) = \mathbf{e}_1 \, F(\mathbf{v}) = \mathbf{e}_1 \, F(\mathbf{e}_1 \, \mathtt{z})[/tex]

    The conventional approach to defining the complex derivative involves the familiar limit:

    [tex]\frac{\partial}{\partial \mathtt{z}} f(\mathtt{z}) = \lim_{\mathit{\tau} \rightarrow 0} \frac{f(\mathtt{z} + \mathit{\tau}) - f(\mathtt{z})}{\mathit{\tau}}[/tex]

    which is said to exist if [itex]f[/itex] satisfies the Cauchy-Riemann condition.

    Using

    [tex]\mathtt{z} = \mathit{x} + i\, \mathit{y} = \mathit{x} + \mathit{y}\, \mathbf{e}_{12}[/tex]

    [tex]\mathtt{z}^\ast = \mathit{x} - i\, \mathit{y} = \mathit{x} - \mathit{y}\, \mathbf{e}_{12}[/tex]

    [tex]\mathit{x} = \frac{\mathtt{z} + \mathtt{z}^\ast}{2}[/tex]

    [tex]\mathit{y}\, \mathbf{e}_{12} = \frac{\mathtt{z} - \mathtt{z}^\ast}{2}[/tex]

    we then have:

    [tex]\frac{\partial}{\partial \mathtt{z}} = \frac{1}{2}(\frac{\partial}{\partial \mathit{x}} - \mathbf{e}_{12} \frac{\partial}{\partial \mathit{y}}) = \frac{1}{2}\nabla \mathbf{e}_1[/tex]

    [tex]\frac{\partial}{\partial \mathtt{z}^\ast} = \frac{1}{2}(\frac{\partial}{\partial \mathit{x}} + \mathbf{e}_{12} \frac{\partial}{\partial \mathit{y}}) = \frac{1}{2}\mathbf{e}_1 \nabla[/tex]

    [tex]\frac{\partial}{\partial \mathit{x}} = \mathbf{e}_1 \cdot \nabla = \frac{\partial}{\partial \mathtt{z}} + \frac{\partial}{\partial \mathtt{z}^\ast} = \frac{1}{2}(\nabla \mathbf{e}_1 + \mathbf{e}_1 \nabla)[/tex]

    And since for analytic [itex]f(\mathtt{z})\;[/itex],

    [tex]\frac{\partial}{\partial \mathtt{z}^\ast} f(\mathtt{z}) = 0\;[/tex],

    [tex]\frac{\partial}{\partial \mathtt{z}} f(\mathtt{z}) = (\frac{\partial}{\partial \mathtt{z}} + \frac{\partial}{\partial \mathtt{z}^\ast}) f(\mathtt{z}) = \frac{\partial}{\partial \mathit{x}} f(\mathtt{z})[/tex]

    Now we can substitute in [itex]F\;[/itex]:

    [tex]\frac{\partial}{\partial \mathtt{z}} f(\mathtt{z}) = \mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} \, F(\mathbf{v}) = \mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} \, F(\mathit{x}\, \mathbf{e}_1 + \mathit{y}\, \mathbf{e}_2)[/tex]

    So we finally arrive at:

    [tex]\frac{\partial}{\partial \mathtt{z}} [\mathbf{e}_1 f(\mathtt{z})] = \lim_{\mathit{\tau} \rightarrow 0} \frac{F(\mathbf{v} + \mathit{\tau}\, \mathbf{e}_1) - F(\mathbf{v})}{\mathit{\tau}}[/tex]

    This is just the directional derivative of [itex]F[/itex] in the direction of [itex]\mathbf{e}_1[/itex] which we could write as:

    [tex]\frac{\partial}{\partial \mathtt{z}} [\mathbf{e}_1 f(\mathtt{z})] = (\mathbf{e}_1 \cdot \nabla) F(\mathit{x}, \mathit{y})[/tex]

    To move back to [itex]C[/itex] we just multiply both sides on the left by [itex]\mathbf{e}_1\;[/itex]:

    [tex]\frac{\partial}{\partial \mathtt{z}} f(\mathtt{z}) = \mathbf{e}_1 (\mathbf{e}_1 \cdot \nabla) F(\mathit{x}, \mathit{y}) = \mathbf{e}_1 \lim_{\mathit{\tau} \rightarrow 0} \frac{F(\mathbf{v} + \mathit{\tau}\, \mathbf{e}_1) - F(\mathbf{v})}{\mathit{\tau}}[/tex]
     
    Last edited: Mar 31, 2006
  10. Apr 2, 2006 #9
    Cauchy-Riemann in R^2

    One last thing regarding Cauchy-Riemann in [itex]R^2\;[/itex].

    We can write the Cauchy-Riemann condition as:

    [tex]\mathbf{e}_1 \nabla f = 0[/tex]

    or left-multiplying both sides by [itex]\mathbf{e}_1[/itex] we get

    [tex]\nabla f = 0[/tex]

    But we saw that [itex]f = \mathbf{e}_1 F\;[/itex].

    So if [itex]f[/itex] is analytic,

    [tex]\nabla f = \mathbf{e}_1 \dot{\nabla} \mathbf{e}_1 \dot{F}[/tex]

    The overdots denote that [itex]\nabla[/itex] acts on [itex]F[/itex] and not just [itex]\mathbf{e}_1\;[/itex]. But we've effectively conjugated the [itex]\nabla[/itex] operator with [itex]\mathbf{e}_1\;[/itex]. Assuming that [itex]\nabla[/itex] operates on everything to its right,

    [tex]\mathbf{e}_1 \nabla \mathbf{e}_1 = \mathbf{e}_1 (\mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} + \mathbf{e}_2 \frac{\partial}{\partial \mathit{y}}) \mathbf{e}_1 = (\frac{\partial}{\partial \mathit{x}} + \mathbf{e}_{12} \frac{\partial}{\partial \mathit{y}}) \mathbf{e}_1 = \mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} - \mathbf{e}_2 \frac{\partial}{\partial \mathit{y}}[/tex]

    So the Cauchy-Riemann condition in [itex]R^2[/itex] can be written as:

    [tex](\mathbf{e}_1 \frac{\partial}{\partial \mathit{x}} - \mathbf{e}_2 \frac{\partial}{\partial \mathit{y}})F = 0[/tex]

    or

    [tex]\mathbf{e}_1 \frac{\partial F}{\partial \mathit{x}} = \mathbf{e}_2 \frac{\partial F}{\partial \mathit{y}}[/tex]

    This is to say that the complex derivative exists if it does not depend on the particular choice we make for the real axis. This makes intuitive sense since spinors do not express any absolute direction, just relative direction. Were the derivative to depend on our choice of real axis then spinors would preferentially act in certain directions on the plane but we've seen how spinors must act on all vectors in the same manner.
     
    Last edited: Apr 2, 2006
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