I am familiar with the Helmholz decomposition of a vector field in three dimensions: $$\vec{V}=\vec{\nabla}\wedge\vec{A}+\vec{\nabla}\phi$$ But I am interested to show that something similar can be obtained for a vector field in higher dimensions. In quantum field theory infact we usually decompose vector fields in longitudinal and transverse components, where the longitudinal component can be written as a gradient and the transverse component is divergenceless.

My question is simple: how this decomposition is done in practice?

The answer I came up with is that one can use the decomposition in symmetric and antisymmetric part of the gradient of the vector field $\nabla_{\mu}A_{\nu}$ to distinguish the longitudinal and transverse components $A^L_{\mu}\;,\;A^T_{\mu}$. That is: \begin{align} \nabla_{\mu}A_{\nu}=\nabla_{(\mu}A_{\nu)}+\nabla_{[\mu}A_{\nu]}\equiv\frac{1}{2}\left(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu}\right)+\frac{1}{2}\left(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}\right) \end{align} and we can always find two vector fields $A^{L}_{\mu}\;,\;A^{T}_{\mu}$ such that \begin{align} \nabla_{\mu}A^L_{\nu}=\nabla_{(\mu}A_{\nu)}\qquad\text{and}\qquad\nabla_{\mu}A^T_{\nu}=\nabla_{[\mu}A_{\nu]} \end{align} which in turn means $\;\nabla_{[\mu}A^L_{\nu]}\,=\,\nabla_{(\mu}A^T_{\nu)}\,\equiv\, 0\;$, so that: \begin{align} \nabla_{\mu}A^L_{\nu}=\nabla_{(\mu}A^L_{\nu)}\qquad\text{and}\qquad\nabla_{\mu}A^T_{\nu}=\nabla_{[\mu}A^T_{\nu]} \end{align} For this reason we can write: \begin{align} \nabla_{\mu}A_{\nu}&=\nabla_{(\mu}A^L_{\nu)}+\nabla_{[\mu}A^T_{\nu]}\equiv\nabla_{(\mu}A^L_{\nu)}+\nabla_{[\mu}A^L_{\nu]}+\nabla_{(\mu}A^T_{\nu)}+\nabla_{[\mu}A^T_{\nu]}\\&=\nabla_{\mu}(A^L_{\nu}+A^T_{\nu}) \end{align} Therefore $A_{\nu}=A^L_{\nu}+A^T_{\nu}+c_{\nu}$, where $c_{\nu}$ is a constant vector. This proves that up to a constant we can decompose the generic vector field $A_{\mu}$ in its longitudinal and transversal components $A^L_{\mu}\;,\;A^T_{\mu}\;$.

By definition follows that the transverse part $A^T_{\mu}$ is divergenceless ($\nabla^{\mu}A^T_{\mu}\,=\,g^{\mu\nu}\nabla_{\mu}A^T_{\nu}\,=\,0$) and the longitudinal part can be written as a gradient ($\nabla_{[\mu}A^L_{\nu]}\,=0\quad\Longrightarrow\;\exists \;\varphi \;\text{so that}\;\;\nabla_{\mu}\varphi=A^L_{\mu}$).

Sot there is a bonus question: is there any problem within this argument? Is there some better way to achieve the decomposition?

  • $\begingroup$ Given $A_b$, is there always a field $A_b^T$ such that $\nabla_a A_b^T=\nabla_{[a} A_{b]}$? $\endgroup$ – Chiral Anomaly Jun 15 '19 at 17:45
  • $\begingroup$ @Chiral mmmm... I'd say yes! maybe the proof must be re-ordered: you show first that you can find a gradient so that $\nabla (A-\nabla\phi)$ is antisymmetric, then you define the transversal part as the difference $A^T:=A-\nabla\phi$ $\endgroup$ – AoZora Jun 15 '19 at 19:33

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