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Let $\mathfrak{u}(n)$ be the Lie algebra of the Lie group $U(n)$. I can define a positive-definite inner product over $\mathfrak{u}(n)$ in this way: if $A,B \in \mathfrak{u}(n)$ I define $\langle A,B \rangle := \Re(\operatorname{Tr}(AB^*))$, where by $\Re$ I denote the real part and by $B^*$ the conjugate transpose of $B$. Why does this inner product over $\mathfrak{u}(n)$ define the unique left-invariant metric on the Lie group $U(n)$?

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    $\begingroup$ For calibration: You're aware that $\langle A, B\rangle$ is (essentially) the Euclidean inner product on $\mathfrak{u}(n)$, and that left multiplication in a Lie group uniquely extends multilinear functions on the Lie algebra to global tensor fields? $\endgroup$ – Andrew D. Hwang Jul 21 '13 at 14:29
  • $\begingroup$ @AndrewD.Hwang Could you specific you comment... I don't understand, sorry. $\endgroup$ – ArthurStuart Jul 21 '13 at 14:34
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Let $A$ and $B$ be complex $n \times n$ matrices with respective $(j, k)$ entries $A_{jk}$ and $B_{jk}$, and note that $B^*$ (the conjugate transpose) has $(j, k)$ entry $\bar{B}_{kj}$. By definition, $$ \langle A, B\rangle = \Re\bigl(\text{Tr}(AB^*)\bigr) = \Re \sum_{j,k=1}^n A_{jk} \bar{B}_{jk},$$ which is precisely the Euclidean inner product of $A$ and $B$ if these matrices are identified with complex vectors in $\mathbf{C}^{n^2}$. The resulting pairing on $\mathfrak{u}(n)$ is the restriction of this inner product.

Generally, if $G$ is a Lie group and $g \in G$, then the left multiplication map $\ell_g:G \to G$ is a diffeomorphism sending $e$ to $g$, so the push-forward $(\ell_g)_*:\mathfrak{g} \to T_gG$ is an isomorphism of vector spaces. An inner product on $\mathfrak{g}$ thereby determines an inner product on each tangent space $T_gG$, and since multiplication is smooth (as a function of $g$) these inner products constitute a Riemannian metric on $G$.

(In case it matters, this left-invariant metric is only "unique" in the sense that it is completely determined by the choice of inner product on $\mathfrak{g}$.)

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Because Tr(AB*) is invariant under transformations of the orthogonal base of the algebra u(n). That's why define invariant metric on group U(n) and is left because B are tensors and must respect the sequence of < A,B >.

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  • $\begingroup$ I'm sorry... but can you be more explicit? $\endgroup$ – ArthurStuart Jul 21 '13 at 15:20

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