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Is $sp(4)$ a subalgebra of $su(5)$? And how can I prove/disprove this?

I know already that it cannot be a regular maximal subgroup of $su(5)$ since the Dynkin diagram (which has two roots of unequal length) cannot be recovered from the (extended) dynkin diagram of $su(5)$. So, if it is a subalgebra, the cartan generators of $sp(4)$ is niet a subset of those of $su(5)$...

Anne O'Nyme
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  • While Qiaochu Yuan's answer ("No") is correct for standard notations, in which $\mathfrak{sp}(4)$ means the compact real form of type $C_{\color{red}{4}}$, it is maybe worthwhile to notice that in the standard notations of split forms, $\mathfrak{sp}4(\mathbb C)$ would e.g. mean the complex simple Lie algebra of type $C{\color{red}{2}}$, which (has dimension $10$ and) is contained in the split form $\mathfrak{sl}_5(\mathbb C)$; actually already in $\mathfrak{sl}_4(\mathbb C)$, i.e. the algebra of type $A_3$. Cf. https://math.stackexchange.com/q/2481396/96384 – Torsten Schoeneberg Jul 23 '19 at 03:41

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No, for dimension reasons. $\dim \mathfrak{sp}(4) = 4 \cdot 9 = 36$ but $\dim \mathfrak{su}(5) = 5^2 - 1 = 24$. The smallest $n$ such that $\mathfrak{sp}(4)$ could embed into $\mathfrak{su}(n)$ on the basis of dimension alone is $n = 7$.

Qiaochu Yuan
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  • Could you explain how you compute the dimensions of $sp(4)$? – Anne O'Nyme Jun 04 '14 at 20:10
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    $\mathfrak{sp}(n)$ is the Lie algebra of $n \times n$ skew-adjoint quaternionic matrices. The entries of such a matrix below the diagonal are uniquely determined by the entries above and there is no restriction on them, so they contribute $4 \frac{n(n-1)}{2} = 2n(n-1)$ to the dimension. The entries on the diagonal must be purely imaginary but there is no additional restriction on them, so they contribute $3n$ to the dimension. So the total dimension is $2n^2 - 2n + 3n = 2n^2 + n = n(2n + 1)$. – Qiaochu Yuan Jun 04 '14 at 20:15
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    Alternatively we can inductively compute the dimension of the corresponding Lie group $\text{Sp}(n)$ as follows. $\text{Sp}(n)$ naturally acts transitively on the unit sphere in $\mathbb{H}^n$, which is $S^{4n-1}$, with stabilizer $\text{Sp}(n-1)$. Hence $\dim \text{Sp}(n) = \dim \text{Sp}(n-1) + 4n-1$, and writing $4n-1 = 2(2n-1) + 1$ you can see by induction that this gives $\dim \text{Sp}(n) = 2n^2 + n$ as above. – Qiaochu Yuan Jun 04 '14 at 20:17