Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:22:06Z http://mathoverflow.net/feeds/question/92624 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivative Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? ssquidd 2012-03-29T23:02:36Z 2012-04-01T08:10:34Z <p>For $\mathbb{C}$-valued functions, why are $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ defined as $$\frac{\partial}{\partial z}= \frac{1}{2}\left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) \quad \quad \frac{\partial}{\partial \bar{z}}= \frac{1}{2}\left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right)$$ where $x$ and $y$ are the real and imaginary parts, respectively.</p> <p>It seems to be me that the inventor of this notation could have easily reverse the signs: Let $\frac{\partial}{\partial z}$ have a $+$ in the middle and let $\frac{\partial}{\partial \bar{z}}$ have a $-$ in the middle instead. Is there any reason that they are defined the way they are now? The same question extends to the $\partial$ and $\bar{\partial}$ operators.</p> <p>One possible guess is that in complex analysis, one usually works with holomorphic functions, so one operator is used much more often than the other. We probably want the more frequently used one to be the one that is easier to write. But I really doubt this is the reason. After all, it's just one extra stroke.</p> http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivative/92634#92634 Answer by Tom Goodwillie for Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? Tom Goodwillie 2012-03-30T00:44:36Z 2012-03-31T13:17:47Z <p>Explaining this notation may be as thankless a task as explaining a joke, but here goes.</p> <p>It may help to think in terms of power series. Consider power series in two variables $x$ and $y$ with complex coefficients, instead of complex functions of real variables $x$ and $y$. Now note that such a thing can be rewritten as a power series in variables $z$ and $\bar z$ by the substitutions $x=\frac{1}{2}(z+\bar z)$ and $y=\frac{1}{2i}(z-\bar z)$. And you can convert back to the other form by the substitutions $z=x+iy$ and $\bar z=x-iy$. The holomorphic case is the case where the only terms in the $(z,\bar z)$ power series are the powers of $z$.</p> <p>Whether you think in terms of power series or not, there is a fruitful fiction that complex functions of $x$ and $y$ can be thought of alternatively as functions of $z$ and $\bar z$, with the holomorphic ones being the functions of $z$ alone. This is consistent with the notation that you are asking about; see Qiaochu's comment to the question.</p> <p>When I was a grad student the story went around that a fellow student attending the complex analysis course questioned precisely these signs, persistently suggesting that the professor had got them wrong. Finally the professor responded "Ah, I see! You and I must be thinking of different square roots of $-1$!"</p> <p>(Edit: That's a joke.)</p> <p>This same fellow student, when we were planning a skit for the annual math department picnic, objected to the title "Let Sleeping Dilogs" on the grounds that people who didn't know about dilogarithms wouldn't get the joke. He wanted to call it "Let Sleeping Dogs Lie" instead, which would have meant that there was no joke to get.</p> http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivative/92641#92641 Answer by Mike Usher for Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? Mike Usher 2012-03-30T03:38:48Z 2012-03-30T03:55:39Z <p>If $f:\mathbb{C}\to\mathbb{C}$ is any smooth function and $z\in\mathbb{C}$, the derivative $df_z$ of $f$ at $z$ is a $\mathbb{R}$-linear operator from the tangent space $T_{z}\mathbb{C}$ to $\mathbb{C}$ (and $T_{z}\mathbb{C}$ can of course be canonically identified with $\mathbb{C}$ since $\mathbb{C}$ is a vector space). Said differently, $df$ is a complex-valued differential $1$-form on $\mathbb{C}$ (i.e. an element of the space $\Omega^1(\mathbb{C};\mathbb{C})=\Omega^1(\mathbb{C})\otimes_{\mathbb{R}} \mathbb{C}$).</p> <p>Any $\mathbb{R}$-linear operator $A$ from one complex vector space to another can be written uniquely as $A=B+C$, where $B$ is complex-linear (i.e. $B(iv)=iBv$ for all $v$) and $C$ is complex anti-linear (i.e. $B(iv)=-iBv$). Specifically, let $Bv=\frac{1}{2}(Av-iAiv)$ and $Cv=\frac{1}{2}(Av+iAiv)$.</p> <p>The operators $\partial$ and $\bar{\partial}$ can then be characterized as follows: for a function $f:\mathbb{C}\to\mathbb{C}$, one has the unique decomposition $$df=\partial f+\bar{\partial} f$$ where the complex-valued $1$-forms $\partial f$ and $\bar{\partial}f$ are such that, at each $z\in\mathbb{C}$, $(\partial f)_z$ is complex-linear and $(\bar{\partial} f)_z$ is complex anti-linear. At least to me that justifies the notation--it makes more sense to have $\partial f$ be the linear part of $df$ and $\bar{\partial} f$ be the antilinear part than the other way around.</p> <p>As for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$, note first that as a special case of the above discussion one has functions $z:\mathbb{C}\to\mathbb{C}$ (which is just the identity) and $\bar{z}:\mathbb{C}\to\mathbb{C}$, and therefore complexified one-forms $dz,d\bar{z}\in\Omega^1(\mathbb{C};\mathbb{C})$. (More specifically, since $z=x+iy$ and $\bar{z}=x-iy$, one has $dz=dx+idy$ and $d\bar{z}=dx-idy$, so for $dz$ and $d\bar{z}$ the signs on $i$ are what you wanted them to be.)</p> <p>At each point of $\mathbb{C}$, $\{dz,d\bar{z}\}$ is a basis (over $\mathbb{C}$) for the complexified cotangent space. The operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ are then naturally defined as the complexified vector fields such that at every point the basis $\{\frac{\partial}{\partial z},\frac{\partial}{\partial\bar{z}}\}$ for the complexified tangent space is the dual basis to the basis $\{dz,d\bar{z}\}$ for the complexified cotangent space. Imposing this dual basis requirement, the formulas $dz=dx+idy$ and $d\bar{z}=dx-idy$ then readily yield the standard formulas for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, and give rise to the pleasant identity, for any smooth $f:\mathbb{C}\to\mathbb{C}$, $$df=\partial f+\bar{\partial} f=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial{\bar{z}}}d\bar{z}$$ just like one would get if $z$ and $\bar{z}$ were real coordinates on a real two-manifold.</p> http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivative/92644#92644 Answer by quasi-coherent for Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? quasi-coherent 2012-03-30T04:06:00Z 2012-03-30T04:06:00Z <p>Note that every $y$ occurs multiplied against $i$, so the non-intrinsicness of $y$ and $i$ cancel out: it is the sign of $iy = (z - \overline{z})/2$ and not of $i$ (or $y$) that is at issue. </p> <p>At the risk of overkill by expressing in another way the same observations which Tom Goodwillie and Mike Usher have made, let's work throughout with $\mathbf{C}$-linear maps and carry out our considerations on a complex manifold $M$ (to get some coordinate-free insight into what is going on). By viewing holomorphic functions as $\mathbf{C}$-valued smooth functions, we can apply $\mathbf{C}$-valued smooth vector fields to holomorphic functions; this is all coordinate-free. For $m \in M$ this defines a canonical $\mathbf{C}$-linear map $$h_m:\mathbf{C} \otimes_{\mathbf{R}} T_m(M') \rightarrow T_m(M)$$ where $M'$ denotes the underlying smooth real manifold. (This map<br> $h_m$ is defined in terms of point derivations at $m$, and is also the tangent map associated to the forgetful morphism of locally ringed spaces $(M, A) \rightarrow (M,O_M)$ over $\mathbf{C}$, where $O_M$ is the sheaf of holomorphic functions and $A$ is the sheaf of $\mathbf{C}$-valued smooth functions.)</p> <p>Now for any $\mathbf{C}$-vector space $V$, we define the "conjugate space" $$\overline{V} := \mathbf{C} \otimes_{\sigma, \mathbf{C}} V$$ where $\sigma(z) = \overline{z}$. For $v \in V$ we denote $1 \otimes v \in \overline{V}$ as $\overline{v}$. Applying scalar extension by complex conjugation to the source and target of $h_m$ and using the $\mathbf{R}$-structure on the source (to identify it with its own conjugate space in a canonical way), we obtain another $\mathbf{C}$-linear map <code>$$h'_m: \mathbf{C} \otimes_{\mathbf{R}} T_m(M') \rightarrow \overline{T_m(M)}.$$</code> Putting the two maps together, we get a $\mathbf{C}$-linear map <code>$$h_m \oplus h'_m: \mathbf{C} \otimes_{\mathbf{R}} T_m(M') \rightarrow T_m(M) \oplus \overline{T_m(M)}.$$</code></p> <p>I claim that this latter map is an isomorphism, and we shall now see this very classical fact by doing a direct computation with local holomorphic coordinates. Writing them in the form $z_k = x_k + iy_k$ for $1 \le k \le n$ (with smooth $\mathbf{R}$-valued functions $x_k$ and $y_k$), one checks that the final displayed map above carries both <code>$\partial_{x_k}|_m$</code> and <code>$(1/i) \otimes \partial_{y_k}|_m$</code> to <code>$(\partial_{z_k}|_m, 0)$</code>, so <code>$$(1/2)(\partial_{x_k}|_m + (1/i) \otimes \partial_{y_k}|_m) \mapsto (\partial_{z_k}|_m, 0)$$</code> Thus, also<br> <code>$$(1/2)(\partial_{x_k}|_m - (1/i) \otimes \partial_{y_k}|_m) \mapsto (0, \overline{\partial_{z_k}|_m}).$$</code> This gives the isomorphism claim, and also "justifies" denoting the respective $\mathbf{C}$-linear derivations <code>$(1/2)(\partial_{x_k} + (1/i)\partial_{y_k})$</code> and <code>$(1/2)(\partial_{x_k} - (1/i)\partial_{y_k})$</code> on the sheaf $A$ of smooth $\mathbf{C}$-valued functions on $M$ as $\partial_{z_k}$ and $\partial_{\overline{z}_k}$.</p> <p>Intrinsically, if $D:A \rightarrow A$ is a $\mathbf{C}$-linear derivation then another is given by $\overline{D}:f \mapsto \overline{D(\overline{f})}$; we may call this the "conjugate" derivation (it is $\mathbf{C}$-linear in $f$!). Obviously <code>$\partial_{\overline{z}_k}$</code> is the conjugate derivation to <code>$\partial_{z_k}$</code> over the coordinate domain, giving a "coordinate-free" link between these two operators.</p> http://mathoverflow.net/questions/92624/why-partial-and-bar-partial-defined-in-that-way-the-wirtinger-derivative/92656#92656 Answer by Peter Dalakov for Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)? Peter Dalakov 2012-03-30T07:59:58Z 2012-04-01T08:10:34Z <p>This is actually a question of linear algebra. The apparently inappropriate choice of sign stems from the fact that if $W$ is a (complex) vector space, and $A$ an endomorphism of $W$, then $$\ker (A-\lambda 1)\subset W$$ is the eigenspace for eigenvalue $(+\lambda)$.</p> <p>In our case, we start with a real vector space, $V$, and an endomorphism $I$, satisfying $I^2=-1$. We extend it $\mathbb{C}$-linearly to an endomorphism $I_{\mathbb{C}}$ of $V\otimes \mathbb{C}$, and decompose the latter into $\pm i$ eigenspaces, usually called $V^{1,0}$ and $V^{0,1}$. The identity transformation decomposes into a sum of (eigenspace) projectors: $$1 = \frac{1}{2}\left( 1 - i I_{\mathbb{C}}\right) + \frac{1}{2}\left( 1 + i I_{\mathbb{C}}\right),$$ and we have a $\mathbb{C}$-linear isomorphism $V\subset V\otimes\mathbb{C}\to V^{1,0} = \ker\frac{1}{2}\left( 1 + i I_{\mathbb{C}}\right)$. Under this isomorphism $V\ni v\mapsto \frac{1}{2}(v-iIv)\in V^{1,0}$. </p> <p>The complex structure $I$ induces a complex structure on the dual space $V^\vee$: this is the dual (transpose) endomorphism: $I^\vee \alpha(v)=\alpha(Iv)$. Its complexification, $I^\vee_\mathbb{C}$, induces an eigenspace decomposition of $V^\vee\otimes\mathbb{C}$.</p> <p>For $\mathbb{R}^2 = (\mathbb{R}^2) ^\vee$ the "standard" complex structure $I$ is represented by the matrix $\left(\begin{array}{rr} 0&amp;-1\\ 1&amp;0\\ \end{array}\right)$, and $I^\vee$ by $\left(\begin{array}{rr} 0&amp;1\\ -1&amp;0\\ \end{array}\right)$. Notice that $I$ is skew-symmetric and the $\pm i$ eigenspaces of $I$ and $I^\vee$ are interchanged! </p> <p>If you apply the above isomorphism $V\simeq V^{1,0}$ ( resp. $V\simeq V^{0,1}$) to the first standard basis vector $e_1\in\mathbb{R}^2$, you will get $$\left\{ \frac{1}{2} \left( \begin{array}{r} 1\\ -i\\ \end{array} \right), \frac{1}{2} \left(\begin{array}{r} 1\\ i\\ \end{array}\right) \right\},$$ an eigenbasis (for $I_\mathbb{C}$) of $\mathbb{C}^2$, where the eigenvalues are ordered $\{+i,-i\}$. Its dual basis is $$\left\{ \left(\begin{array}{r} 1\\ i\\ \end{array} \right), \left(\begin{array}{r} 1\\ -i\\ \end{array} \right) \right\},$$ consisting of eigenvectors for $I^\vee_\mathbb{C}$, with eigenvalues $\{+i,-i\}$.</p> <p>Now rephrase all of the above in terms derivations. The $\mathbb{C}$-isomorphism $V\simeq V^{1,0}$ gives you $\frac{\partial}{\partial x}\mapsto \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)$, and $$\left\{ \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) \right\}$$ is a $\mathbb{C}$-basis of $\mathbb{R}^2\otimes \mathbb{C}$, dual to $\left\{dx + i dy, dx-idy \right\}$. As the latter is conventionally denoted by $\left\{dz,d\overline{z}\right\}$, it is natural to denote its dual basis by $\left\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\right\}$.</p> <p>Notice that with this convention $dz\left(\frac{\partial}{\partial \overline{z}} \right) =0$ and $d\overline{z}\left(\frac{\partial}{\partial \overline{z}} \right) =1$, which is highly desirable, as mentioned in the comments.</p> <p><strong>ADDENDUM</strong></p> <p>In sum, we have to make a compatible choice of:<p></p> <p>1) Eigenvectors of $I$<br> 2) Eigenvectors of $I^\vee$.<br></p> <p>We tend to put bars on the eigenvectors with eigenvalue $(-i)$. Eigenvectors are determined up to a nonzero scalar, so <em>a priori</em> we have to choose 4 of these (in $\mathbb{C}^2=\mathbb{R}^2\otimes\mathbb{C}$ ). However, the conditions:<p> a) $\frac{\partial}{\partial \overline{z}} = \overline{\frac{\partial}{\partial z}}$<br> b)$\left\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\right\}$ is a dual $\mathbb{C}$-basis to $\left\{dz,d\overline{z}\right\}$<br> leave us only with the freedom of replacing $\frac{\partial}{\partial z}$ by a scalar multiple. </p> <p>As an example, if you take the complex structure $\left(\begin{array}{rr} -1&amp;-2\\ 1&amp;1\\ \end{array}\right)$, then, up to that single scalar ambiguity, we have $$dz = dx+ (1+i)dy, \ d\overline{z} = dx + (1-i) dy$$ $$\frac{\partial}{\partial z}=\frac{1}{2}\left( (1+i)\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left( (1-i)\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)$$</p>