# Connections, The Kähler Condition, and Curvature – III. The Kähler Condition

Suppose ${M}$ is a complex manifold of complex dimension ${n}$. We can regard ${M}$ as a real manifold of real dimension ${m = 2n}$. For local holomoprhic coordinates ${z^1, \dots, z^n}$, we write ${z^\alpha = x^\alpha + \sqrt{-1}y^\alpha = x^\alpha + \sqrt{-1}x^{n + \alpha}}$. Then ${{\partial\over{\partial x^\alpha}} = 2\,\text{Re}\,{\partial\over{\partial z^\alpha}}}$ and ${{\partial\over{\partial y^\alpha}} = 2\,\text{Re}\,\sqrt{-1}{\partial\over{\partial z^\alpha}}}$. The complexification ${T_M \otimes \mathbb{C}}$ of ${T_M}$ splits into a direct sum ${T_M^{1,0} \oplus T_M^{0,1}}$, with ${T_M^{1,0}}$ spanned by ${{\partial\over{\partial z^\alpha}}}$ and ${T_M^{0,1}}$ spanned by ${{\partial\over{\partial\overline{z^\alpha}}}}$. We have a natural map from ${T_M}$ to ${T_M^{0, 1}}$, which is the composite of the inclusion map ${T_M \rightarrow T_M \otimes \mathbb{C}}$ and the projection map ${T_M \otimes \mathbb{C} = T_M^{1, 0} \oplus T_M^{0, 1} \rightarrow T_M^{1, 0}}$ onto the first summand. This natural map is an isomorphism, and its inverse is the map ${2\,\text{Re}\,(\cdot)}$ from ${T_M^{0, 1}}$ to ${T_M}$. Through the ${\mathbb{R}}$-isomorphism ${2\,\text{Re}\,(\cdot)}$, the operator of multiplication given by ${\sqrt{-1}}$ in ${T_M^{1, 0}}$ corresponds to some operator ${J}$ in ${T_M}$. From ${{\partial\over{\partial x^\alpha}} = 2\,\text{Re}\,{\partial\over{\partial z^\alpha}}}$ and ${{\partial\over{\partial y^\alpha}} = 2\,\text{Re}\,\sqrt{-1}{\partial\over{\partial z^\alpha}}}$, it follows that the operator ${J: T_M \rightarrow T_M}$ is given by ${J({\partial\over{\partial x^\alpha}}) = {\partial\over{\partial y^\alpha}}}$ and ${J({\partial\over{\partial y^\alpha}}) = -{\partial\over{\partial x^\alpha}}}$. This means that if ${T_M}$ is made into a ${\mathbb{C}}$-vector space by defining multiplication by ${\sqrt{-1}}$ as ${J}$, then ${2\,\text{Re}\,(\cdot)}$ is a ${\mathbb{C}}$-isomorphism between the ${\mathbb{C}}$-vector spaces ${T_M^{1, 0}}$ and ${T_M}$. We identify ${T_M^{1, 0}}$ and ${T_M}$ through this isomorphism. Note that

$\displaystyle J\left({\partial\over{\partial z^\alpha}}\right) = J\left({1\over2}\left({\partial\over{\partial x^\alpha}} - \sqrt{-1}{\partial\over{\partial y^\alpha}}\right)\right) = {1\over2}\left({\partial\over{\partial y^\alpha}} + \sqrt{-1}{\partial\over{\partial x^\alpha}}\right) = \sqrt{-1}{\partial\over{\partial x^\alpha}},$

and by analogous direct derivation or from the fact that ${J}$ is real, one has ${J({\partial\over{\partial \overline{z^\alpha}}}) = -\sqrt{-1}{\partial\over{\partial \overline{z^\alpha}}}}$. This simply means that ${T_M^{1, 0}}$ and ${T_M^{0, 1}}$ are respectively the eigenspaces of ${J}$ for the eigenvalues ${\sqrt{-1}}$ and ${-\sqrt{-1}}$.

Since ${T_M}$ with the operator ${J}$ becomes a ${\mathbb{C}}$-vector space, when we have a metric on ${T_M}$ to measure the length of the vectors on ${T_M}$, we can ask whether this length function comes from a Hermitian inner product. One necessary condition is that the length of a vector should be invariant under ${J}$. It turns out that this condition is also sufficient. So we are going to use this condition as definition of a Hermitian metric and then verify that it agrees with the usual notion of a Hermitian metric. We say that a Riemannian metric ${g}$ on ${T_M}$ is Hermitian if ${g(JX, JY) = g(X, Y)}$ for any ${X}$, ${Y}$ in ${T_M}$. Now we are going to reconcile this definition with the usual definition of a Hermitian metric respresented by a Hermitian matrix. A Riemannian metric is a real-valued ${\mathbb{R}}$-billinear function on ${T_M \times T_M}$. We can extend it by ${\mathbb{C}}$-bilinearity to a ${\mathbb{C}}$-bilinear function on ${(T_M \otimes \mathbb{C}) \times (T_M \otimes \mathbb{C})}$. Then the Riemannian metric ${g}$ is broken into four parts ${g_{\alpha\beta}}$, ${g_{\alpha\overline{\beta}}}$, ${g_{\overline{\alpha}\beta}}$, ${g_{\beta\alpha\overline{\beta}}}$. We investigate the condition ${g(JX, JY) = g(X, Y)}$ for each part. For the part ${g_{\alpha\beta}}$, since ${J|T_{M}^{1, 0}}$ is the same as multiplication by ${\sqrt{-1}}$, we have ${g(\sqrt{-1}X^{1, 0},\sqrt{-1}X^{1, 0}) = g(X^{1, 0}, X^{1, 0})}$, and by ${\mathbb{C}}$-bilinearity, we have ${g_{\alpha\beta} = 0}$. From an analogous argument or from the fact that ${g}$ is real, we have ${g_{\overline{\alpha}\overline{\beta}} = 0}$.

Since ${g}$ is symmetric, we have ${g_{\alpha\overline{\beta}} = g_{\overline{\beta}\alpha}}$. Moreover, since ${g}$ is real, we must have ${\overline{g({\partial\over{\partial z^\alpha}}, {\partial\over{\partial \overline{z^\beta}}})} = g({\partial\over{\partial \overline{z^\alpha}}}, {\partial\over{\partial z^\beta}})}$, i.e. ${\overline{g_{\alpha\overline{\beta}}} = g_{\overline{\alpha}\beta}}$. Hence ${\overline{g_{\alpha\overline{\beta}}} = g_{\beta\overline{\alpha}}}$, and ${g_{\alpha\overline{\beta}}}$ is a Hermitian matrix. So the condition ${g(JX, JY) = g(X, Y)}$ simply means that ${g_{\alpha\beta} = g_{\overline{\alpha}\overline{\beta}} = 0}$, ${g_{\alpha\overline{\beta}}}$ is Hermitian, and ${g_{\overline{\beta}\alpha}}$ is the transpose of ${g_{\alpha\overline{\beta}}}$.

The correspondence ${2\,\text{Re}\,(\cdot)}$ between ${T_M^{1, 0}}$ and ${T_M}$ transports the metric ${g}$ of ${T_M}$ to a metric on ${T_M^{1, 0}}$, and we want to determine this metric on ${T_M}$. For this metric on ${T_M^{1, 0}}$, the inner product of ${{\partial\over{\partial z^\alpha}}}$ and ${{\partial\over{\partial z^\beta}}}$ is

$\displaystyle g\left({\partial\over{\partial z^\alpha}} + \overline{\partial\over{\partial z^\alpha}}, {\partial\over{\partial z^\beta}} + \overline{\partial\over{\partial z^\beta}}\right) = g_{\alpha\overline{\beta}} + g_{\overline{\alpha}\beta} = g_{\alpha\overline{\beta}} + g_{\beta\overline{\alpha}} = 2\,\text{Re}\,g_{\alpha\overline{\beta}}.$

So the Riemannian metric ${g}$ on ${T_M}$ correpsonds to the metric ${2\,\text{Re}(g_{\alpha\overline{\beta}}dz^\alpha d\overline{z^\beta})}$ on ${T_M^{1, 0}}$. If we only look at the length of a vector, the metric on ${T_M^{1, 0}}$ is simply the Hermitian metric ${2g_{\alpha\overline{\beta}}dz^\alpha d\overline{z^\beta}}$.Let us now explain the difference between ${2\text{Re}\,g_{\alpha\overline{\beta}}}$ and ${2g_{\alpha\overline{\beta}}}$. A metric is used to measure the length of a vector. In that sense, ${2\text{Re}\,g_{\alpha\overline{\beta}}}$ and ${2g_{\alpha\overline{\beta}}}$ agree. The difference comes in when we want to express the square of the length of the vector as its inner product with itself. In the case of an ${\mathbb{R}}$-vector space, the inner product should be an ${\mathbb{R}}$-bilinear functional. In the case of a ${\mathbb{C}}$-vector space, the inner product should be a functional which is ${\mathbb{C}}$-linear in the first argument and ${\mathbb{C}}$-conjugate linear in the second argument. When the lengths of the vectors are determined, in either case there is only one way of writing out such an inner product, and it is done by polarization. When we regard ${T_M^{1, 0}}$ has an ${\mathbb{R}}$-vector space, we have the inner product ${2\,\text{Re}\,g_{\alpha\overline{\beta}}}$ for the metric, whereas when we regard ${T_M^{1, 0}}$ as a ${\mathbb{C}}$-vector space, we have the Hermitian inner product ${2g_{\alpha\overline{\beta}}}$ for the metric.

If one wants to keep working only with real vectors all the time, one can verify directly in the following way that the condition ${g(JX, JY) = g(X, Y)}$ implies that the length function defined by ${g}$ comes from a Hermitian inner product on ${T_M}$ when ${T_M}$ is made into a ${\mathbb{C}}$-vector space by the operator ${J}$. The map ${2\,\text{Re}\,(\cdot)}$ is a ${\mathbb{C}}$-isomorphism between ${T_M^{1, 0}}$ and ${T_M}$. The inverse of this map is ${X \rightarrow {1\over2}(X - iJX)}$ because ${J({\partial\over{\partial x^\alpha}}) = {\partial\over{\partial y^\alpha}}}$. Hence we expect the Hermitian inner product to be ${2g({1\over2}(X - iJX), {1\over2}(Y + iJY))}$ when ${g}$ is extended by ${\mathbb{C}}$-bilinearity. In other words, the Hermitian inner product is

$\displaystyle {1\over2}g(X, Y) + {i\over2}g(X, JY) - {i\over2}g(JX, Y) + {1\over2}g(JX, JY),$

which is the same as ${g(X, Y) + ig(X, JY)}$. This procedure is the same as getting the Hermitian inner product from the square of the length by polarization. We would like to note that ${g(X, Y)}$ is the real part of the Hermitian inner product, and ${g(X, JY)}$ is its imaginary part. In terms of abstract algebra, the above argument can be formulated in the following way. An ${\mathbb{R}}$-bilinear inner product on a ${\mathbb{C}}$-vector space is the real part of a Hermitian inner product if and only if it is invariant under the operation of multiplication by ${\sqrt{-1}}$.

Let us now consider the Levi-Civita connection expressed in terms of the complex coordinate indices ${\alpha}$, ${\overline{\alpha}}$. We let ${i}$ denote both ${\alpha}$ and ${\overline{\alpha}}$. We have

$\displaystyle \Gamma_{jk}^i = {1\over2}g^{i\ell}(\partial_j g_{k\ell} + \partial_k g_{j\ell} - \partial_\ell g_{jk}).$

Suppose ${i = \alpha}$. Then ${\ell}$ must be of type ${\overline{\lambda}}$ to make a nonzero contribution. So

$\displaystyle \Gamma_{\beta\overline{\gamma}}^\alpha = {1\over2}g^{\alpha\overline{\lambda}}(\partial_\beta g_{\overline{\gamma}\overline{\lambda}} + \partial_{\overline{\gamma}} g_{\beta\overline{\lambda}} - \partial_{\overline{\lambda}}g_{\beta\overline{\gamma}}) = {1\over2}g^{\alpha\overline{\lambda}}(\partial_{\overline{\gamma}}g_{\beta\overline{\lambda}} - \partial_{\overline{\lambda}}g_{\beta\overline{\gamma}})$

and

$\displaystyle \Gamma_{\overline{\beta}\overline{\gamma}}^\alpha = {1\over2}g^{\alpha\overline{\lambda}}(\partial_{\overline{\beta}}g_{\overline{\gamma}\overline{\lambda}} + \partial_{\overline{\gamma}}g_{\overline{\beta}\overline{\gamma}} - \partial_{\overline{\lambda}}g_{\overline{\beta}\overline{\gamma}}) = 0.$

These Christoffel symbols are the connection of ${T_M \otimes \mathbb{C}}$, which is induced by the Levi-Civita connection of ${T_M}$. In general, this connection of ${T_M \otimes \mathbb{C}}$ does not define a connection of ${T_M^{1, 0}}$, because ${\Gamma_{\beta\overline{\gamma}}^{\overline{\alpha}} = \overline{\Gamma_{\gamma\overline{\beta}}^\alpha}}$ may not be zero. So when we differentiate a section of ${T_M^{1, 0}}$ in the ${(0, 1)}$ direction, we may get a section of ${T_M \otimes \mathbb{C}}$ which is not entirely in ${T_M^{1, 0}}$. We get a connection of ${T_M^{1, 0}}$ if and only if ${\Gamma_{\beta\overline{\gamma}}^\alpha}$ vanishes.

We want to get a connection of ${T_M^{1, 0}}$ from ${T_M}$ through the ${\mathbb{R}}$-isomorphism ${2\,\text{Re}\,(\cdot)}$. In the correspondence ${2\,\text{Re}\,(\cdot)}$ between ${T_M^{1, 0}}$ and ${T_M}$, when we take a section ${s}$ of ${T_M^{1, 0}}$, we should consider ${D(s + \overline{s})}$ in ${T_M}$, which must be of the form ${t + \overline{t}}$. Then ${Ds}$ in ${T_M^{1, 0}}$ is equal to ${t}$. So when we take ${D{\partial\over{\partial z^\alpha}}}$, we should be looking at

$\displaystyle D\left({\partial\over{\partial z^\alpha}} + \overline{\partial\over{\partial z^\alpha}} \right) = (\Gamma_{\alpha j}^i + \Gamma_{\overline{\alpha}j}^i){\partial\over{\partial z^i}} \otimes dz^j$

$\displaystyle =(\Gamma_{\alpha\gamma}^\beta + \Gamma_{\overline{\alpha}\gamma}^\beta){\partial\over{\partial z^\beta}} \otimes dz^\gamma + (\Gamma_{\alpha\overline{\gamma}}^\beta + \Gamma_{\overline{\alpha}\overline{\gamma}}^\beta){\partial\over{\partial z^\beta}} \otimes dz^{\overline{\gamma}}$

$\displaystyle +(\Gamma_{\alpha\gamma}^{\overline{\beta}} + \Gamma_{\overline{\alpha}\gamma}^{\overline{\beta}}){\partial\over{\partial z^{\overline{\beta}}}} \otimes dz^\gamma + (\Gamma_{\alpha\overline{\gamma}}^{\overline{\beta}} + \Gamma_{\overline{\alpha}\overline{\gamma}}^{\overline{\beta}}){\partial\over{\partial z^{\overline{\beta}}}} \otimes dz^{\overline{\gamma}}$

$\displaystyle =(\Gamma_{\alpha\gamma}^\beta + \Gamma_{\overline{\alpha}\gamma}^\beta){\partial\over{\partial z^\beta}} \otimes dz^\gamma + \Gamma_{\alpha\overline{\gamma}}^\beta {\partial\over{\partial z^\beta}} \otimes dz^{\overline{\gamma}}$

$\displaystyle +\Gamma_{\overline{\alpha}\gamma}^{\overline{\beta}}{\partial\over{\partial z^{\overline{\beta}}}} \otimes dz^\gamma + (\Gamma_{\alpha\overline{\gamma}}^{\overline{\beta}} + \Gamma_{\overline{\alpha}\overline{\gamma}}^{\overline{\beta}}){\partial\over{\partial z^{\overline{\beta}}}} \otimes dz^{\overline{\gamma}}.$

Here ${i}$, ${j}$ go through the range ${1, \dots, n, \overline{1}, \dots, \overline{n}}$. Thus in ${T_M^{1, 0}}$, we have

$\displaystyle D\left({\partial\over{\partial z^\alpha}}\right) = (\Gamma_{\alpha\gamma}^\beta + \Gamma_{\overline{\alpha}\gamma}^\beta){\partial\over{\partial z^\beta}} \otimes dz^\gamma + \Gamma_{\alpha\overline{\gamma}}^\beta {\partial\over{\partial z^\beta}} \otimes dz^{\overline{\gamma}}.$

So the connection is complex if and only if ${\Gamma_{\alpha\overline{\gamma}}^\beta = 0}$ The induced connection of ${T_M^{1, 0}}$ is automatially compatible with the metric because the Levi-Civita connection of ${T_M}$ is compatible with the metric and the length functions of the metrics of ${T_M^{1, 0}}$ and ${T_M}$ correspond through ${2\,\text{Re}\,(\cdot)}$. Moreover, when the connection is complex, the connection is given by ${\omega_\alpha^\beta = \Gamma_{\alpha\gamma}^\beta dz^\gamma}$. The formula for the Christoffel symbol ${\Gamma_{\alpha\gamma}^\beta}$ of the Levi-Civita connection gives

$\displaystyle \Gamma_{\alpha\beta}^\alpha = {1\over2}g^{\alpha\overline{\lambda}} (\partial_\beta g_{\lambda\overline{\lambda}} + \partial_\gamma g_{\beta\gamma} - \partial_{\overline{\lambda}}g_{\beta\gamma}) = {1\over2}g^{\alpha\overline{\lambda}}(\partial_\gamma g_{\beta\gamma} - \partial_{\overline{\lambda}}g_{\beta\gamma}) = g^{\alpha\overline{\lambda}}\partial_\beta g_{\gamma\overline{\lambda}}.$

We note that this connection is the same as the complex metric connection for the holomorphic vector bundle ${T_M^{1, 0}}$ with the Hermitian metric ${g_{\alpha\overline{\beta}}}$ (or ${2g_{\alpha\overline{\beta}}}$). The Hermitian metric is called Kähler if the Levi-Civita connection is complex. An equivalent condition is that ${\partial_{\overline{\gamma}}g_{\beta\overline{\lambda}} = \partial_{\overline{\lambda}}g_{\beta\overline{\gamma}}}$. This is equivalent to the condition that the ${2}$-form ${g_{\alpha\overline{\beta}}dz^\alpha \wedge d\overline{z^\beta}}$ is closed. The ${2}$-form ${2\sqrt{-1}g_{\alpha\overline{\beta}}dz^\alpha \wedge d{\overline{z^\beta}}}$ is equal to

$\displaystyle \sqrt{-1}g_{\alpha\overline{\beta}}dz^\alpha \otimes d\overline{z^\beta} - \sqrt{-1}g_{\alpha\overline{\beta}}d\overline{z^\beta} \otimes dz^\alpha$

$\displaystyle =-(g_{\alpha\overline{\beta}}dz^\alpha \otimes d\overline{\sqrt{-1}z^\beta} + g_{\alpha\overline{\beta}}d\overline{z^\beta} \otimes \sqrt{-1}dz^\alpha).$

Thus

$\displaystyle 2\sqrt{-1}g_{\alpha\overline{\beta}}dz^\alpha \wedge d\overline{z^\beta}(X, Y) = -g(X, JY),$

because ${g}$ is ${2\,\text{Re}(g_{\alpha\overline{\beta}}dz^\alpha \otimes d\overline{z^\beta})}$ and ${dz^\beta(JY) = \sqrt{-1}dz^\beta(Y)}$. The function ${g(X, JY)}$ is skew-symmetric in ${X}$ and ${Y}$ because ${g(JX, JY) = g(X, Y)}$. So ${g(X, JY)}$ defines a ${2}$-form on ${M}$. Consider the Hermitian form ${g_{\alpha\overline{\beta}}dz^\alpha \otimes d{\overline{z^\beta}}}$. Twice its imaginary part is ${-\sqrt{-1}(g_{\alpha\overline{\beta}}dz^\alpha \otimes d{\overline{z^\beta}} - g_{\overline{\alpha}\beta}d\overline{z^\alpha} \otimes dz^\beta)}$, which is ${-2\sqrt{-1}g_{\alpha\overline{\beta}}dz^\alpha \wedge d\overline{z^\beta}}$. Thus the ${2}$-form ${g(X, JX)}$ is the imaginary part of ${g_{\alpha\overline{\beta}}dz^\alpha \otimes d\overline{z^\beta}}$, and the Riemannian metric is twice its real part. We have also seen this before. The ${2}$-form is called the fundamental form of the Hermitian metric. So a Hermitian metric is Kähler if and only if its fundamental form is closed.

We now look at another characterization of the Kähler condition for a Hermitian metric. This characterization is that for any point ${P}$ of ${M}$, there exists a local holomorphic coordinate system ${z^\alpha}$ centered at ${P}$ so that the first derivative of ${g_{\alpha\overline{\beta}}}$ vanishes at ${P}$. Clearly if such a coordinate system exists, then the fundamental form of the Hermitian metric is zero. Conversely suppose the metric is Kähler. Without loss of generality, we can assume that ${g_{\alpha\overline{\beta}}}$ equals the Kronecker delta at the point ${P}$. We seek a new coordinate system ${w^\alpha}$ which is related to ${z^\alpha}$ by ${z^\alpha = w^\alpha + c_{\beta\gamma}^\alpha w^\beta w^\gamma}$, with ${c_{\beta\gamma}^\alpha}$ symmetric in ${\beta}$ and ${\gamma}$. Let ${h_{\alpha\overline{\beta}}dw^\alpha d\overline{w^\beta} = g_{\alpha\overline{\beta}}dz^\alpha d\overline{z^\beta}}$. Since ${c_{\beta\gamma}^\alpha}$ is symmetric in ${\beta}$ and ${\gamma}$, it follows that

$\displaystyle h_{\gamma\overline{\mu}} = g_{\alpha\overline{\beta}} (\delta_\lambda^\alpha + 2c_{\lambda\sigma}^\alpha w^\sigma)\overline{(\delta_\mu^\beta + 2c_{\mu\tau}^\beta w^\tau)},$

and ${dh_{\lambda\overline{\mu}}}$ at ${P}$ equals

$\displaystyle dg_{\lambda\overline{\mu}} + 2c_{\lambda \sigma}^\mu dw^\sigma + 2\overline{c_{\mu\tau}^\lambda}dw^\tau.$

Thus we should choose ${c_{\lambda\sigma}^\mu = -{1\over2}\partial_\sigma g_{\lambda\overline{\mu}}}$. This can be done with ${c_{\lambda\sigma}^\mu}$ symmetric in ${\lambda}$ and ${\sigma}$ if and only if ${\partial_\sigma g_{\lambda\overline{\mu}}}$ is symmetric in ${\lambda}$ and ${\mu}$.

We would like to remark why in the real case one can always find local coordinates so that the first derivative of the coefficients of the metric tensor ${g_{ij}}$ vanishes at a point. In the real case, we want ${dg_{ij} + 2c_{ik}^jdw^k + 2c_{jk}^idw^k}$ to vanish at ${P}$. In other words, ${c_{ik}^j + c_{jk}^i = -{1\over2}\partial_kg_{ij}}$ with ${c_{jk}^i}$ symmetric in ${j}$ and ${k}$. Recall the equations ${\partial_k g_{ij} = \Gamma_{j, ik} + \Gamma_{i, jk}}$ which were used to solve for the Christoffel symbols in terms of the Riemannian metric. All we have to do is set ${c_{jk}^i = -{1\over2} \Gamma_{i, jk}}$. This solution is the same as getting a coordinate system by integrating out along geodesics emanating from ${P}$.

Another characterization of the Kähler condition for a Hermitian metric is that all the Christoffel symbols ${\Gamma_{jk}^i}$ for the Levi-Civita connection vanish except the types ${\Gamma_{\beta\gamma}^\alpha}$ and its complex conjugate ${\Gamma_{\overline{\beta}\overline{\gamma}}^{\overline{\alpha}}}$. We have seen that ${\Gamma_{\overline{\beta}\overline{\gamma}}^{{\alpha}}}$ always vanishes. We have the vanishing of ${\Gamma_{{\beta}\overline{\gamma}}^{{\alpha}}}$ if and only if the Hermitian metric is Kähler. We get our conclusion because ${\Gamma_{jk}^i}$ is symmetric in ${j}$ and ${k}$, and it is real in the sense that ${\overline{\Gamma_{jk}^i} = \Gamma_{\overline{j}\overline{k}}^{\overline{i}}}$. Geometrically, this characterization of a Kähler metric says that a Hermitian metric is Kähler if and only if the covariant derivative of a vector field of type ${(1, 0)}$ is still of type ${(1, 0)}$. In other words, types are preserved under parallel transport. This is equivalent to our earlier observation that ${\Gamma_{\beta\overline{\gamma}}^\alpha}$ vanishes if and only if the connection of ${T_M \otimes \mathbb{C}}$ induced from the Levi-Civita connection of ${T_M}$ defines a connection of ${T_M^{1, 0}}$ through the inclusion map ${T_M^{1, 0} \rightarrow T_M \otimes \mathbb{C}}$.