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I am trying to derive equation (2.4.2) in Polchinski's string theory textbook,

$$\overline \partial T_{zz}=\partial T_{\overline z \overline z} = 0 \tag{2.4.2}.$$

Using the conservation equation, tracelessness, and symmetry, I've found that,

$T_{zz} = \frac{1}{2}(T^{11} - iT^{12}), \qquad T_{\overline z \overline z} = \frac{1}{2}(T^{11} + iT^{12})$;

$\overline \partial T_{zz} = \frac{1}{4}(\partial_1 T_{11} + \partial_2 T^{21})(1+i), \qquad \partial T_{\overline z\overline z} = \frac{1}{4}(\partial_1 T_{11} + \partial_2 T^{21})(1-i)$;

$\overline \partial T_{zz} = i \partial T_{\overline z \overline z} $.

If I write out the conservation equation explicitly,

$\partial_a T^{ab} = \partial_1 T^{11} + \partial_1 T^{12} + \partial_2 T^{21} + \partial_2 T^{22}$,

and rewrite the partial derivatives and energy momentum tensor components as combinations of the complex variables, I have found that it is equivalent to

$ \partial T_{\overline z \overline z} + \overline \partial T_ {zz} + i(\overline \partial T_{zz} - \partial T_{\overline z\overline z}) = 0$

But I don't see how this is implies (2.4.2). The term in parentheses is already imaginary so it won't vanish seperately from the first two terms.

Any help would be appreciated, thanks.

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$\partial_b T^{ba} = 0$ is two sets of equations - one for $a = 1$ and another for $a = 2$. The sum is only done over the $b$ index. – Prahar May 3 at 14:47

1 Answer 1

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$\partial_a T^{ab}=0$ implies $\partial_1 T^{11}+\partial_2 T^{21}=0$. You don't sum over the uncontracted index $b$.

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Of course! Thank you, I knew I was missing something obvious. – Haz May 4 at 0:52

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