Is there any geometric or more direct conceptual way to understand a supersymmetry algebra, rather than starting from a Lagrangian including boson and fermion fields, deriving all the expressions ensuring the supersymmetry invariance and then writing down the supersymmetry algebra? For a geometric or algebraic way, I mean to (at least partially) derive the supersymmetry algebra from pure geometry (spinors, spin group, etc.).

1$\begingroup$ Could you clarify? Do you just want SuperPoincare(4)? Other dimensions? Higher $\mathcal{N}$? $\endgroup$ – AHusain Aug 1 '17 at 21:01

$\begingroup$ any dimension, any N. $\endgroup$ – Hao Yu Aug 1 '17 at 21:36

$\begingroup$ By Deligne's theorem on tensor categories, the group of symmetries is at least a supergroup ncatlab.org/nlab/show/… $\endgroup$ – user40276 Aug 14 '17 at 1:51
In this paper about a global theory of supermanifolds, Alice Rogers develops the theory of supermanifolds as they underly supersymmetric field theories from a rigorous but physicistfriendly differentialgeometric point of view from topological scratch, and also puts some additional structures such as vector fields and tangent spaces on them. She also compares the $G^{\infty}$ or deWitt supermanifolds to the algebrogeometric approach of for example Kostant or Leites.
Alice Roger's 2007 textbook explains the supermathematics needed for doing superphysics (including Grassmann algebras, super Lie groups such as the super Poincare group, etc) and contains more applications to different physics topics such as $N=1$ supersymmetry, supergravity, some aspects of string theory, or Brownian motion from the same nice differentialgeometric point of view.
I'm not sure what you mean by "derive".
For a more mathematical and geometric description of the super Poincaré group in general dimension you could check out
 Freed, Lectures on field theory and supersymmetry (Lecture 6);
 Freed, Five lectures on supersymmetry, AMS 1999 (Lecture 3);
 Deligne and Freed, Supersolutions [arXiv:hepth/9901094] (Section 1.1), which can also be found in Quantum fields and strings: a course for mathematicians;
plus the three additional references given for superPoincaré group at nLab.
For the 4d case see also
 Costello, e.g. [arXiv:1401.2676] (Section 1.1).

1
If you are looking for geometricalgebraic interpretations of supersymmetric field theories, then noncommutative geometry in the sense of A. Connes seems to be the natural playground. There has been quite a lot of literature on the subject since the '80's. Just to mention some inspiring works:
 Supersymmetry and NonCommutative Geometry, J. Fröhlich, O. Grandjean, A. Recknagel, Quantum Fields and Quantum Space Time, 1997
 Supersymmetry and noncommutative geometry, W.Kalau, M.Walzeb, Journal of Geometry and Physics, v.22, 1, 1997, 77102
 Noncommutative geometry and string field theory, E. Witten, Nuclear Physics B, v.268, 2, 1986, 253294
See also this, which is a more modern work, in the form of a book, discussing (noncommutative) geometrical ideas related to supersymmetry although from a more imo phenomenological viewpoint (emphasizing the supersymmetric standard model, supersymmetry breaking, elementary particle interactions , etc).
Edit:
After giving some further thinking on your question and since you are asking for some method
... to derive the supersymmetry algebra from pure geometry ...
i decided to try to provide an example of such a construction. This will still be in the spirit of the noncommutative geometrical approach but in a somewhat simplified manner:
Example: Consider the (usual) complex variables $x_{i}$ and the anticommuting Grassman variables $\xi_{j}$. (The $x_{i}$ commute with each other and with the $\xi_{j}$ as well, while the $\xi_{j}$ anticommute with each other). Next consider the space of polynomials in those variables and their formal derivatives. If we use the notation: $$ \begin{array}{cccc} \begin{array}{ccc} b_{i}^{+} & \leftrightarrow & x_{i} \\ b_{i}^{} & \leftrightarrow & \frac{\partial}{\partial{x_{i}}} \end{array} & & & \begin{array}{ccc} f_{i}^{+} & \leftrightarrow & \xi_{i} \\ f_{i}^{} & \leftrightarrow & \frac{\partial}{\partial{\xi_{i}}} \end{array} \end{array} $$ it is easy to see that the multiplication and the differential operators $(x_{i},\frac{\partial}{\partial{x_{i}}})$ acting on (usual) complex variables correspond to boson creationannihilation operators and that the multiplication and the differential operators $(\xi_{i},\frac{\partial}{\partial{\xi_{i}}})$ acting on Grassmann (anticommuting) variables correspond to fermion creationannihilation operators, in the sense that the following relations hold: $$ \begin{array}{c} [b_{i}^{}, b_{j}^{+}]\equiv b_{i}^{}b_{j}^{+}  b_{j}^{+}b_{i}^{} = \delta_{ij} I, \ \ \ \ \ \ [b_{i}^{}, b_{j}^{}] = [b_{i}^{+}, b_{j}^{+}] = 0 \ \ \ \ \ \ \ \ \ (1) \\ \\ \{f_{i}^{}, f_{j}^{+}\}\equiv f_{i}^{}f_{j}^{+} + f_{j}^{+}f_{i}^{} = \delta_{ij} I, \ \ \ \ \ \ \{f_{i}^{}, f_{j}^{}\} = \{f_{i}^{+}, f_{j}^{+}\} = 0 \ \ \ \ \ \ \ \ (2) \\ \\ f_{j}^{\varepsilon}b_{i}^{\eta} = b_{i}^{\eta}f_{j}^{\varepsilon} \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \end{array} $$ where $\varepsilon,\eta=\pm$ and $I$ is the identity operator. This is an algebra defined in terms of generators and relations and it is a quotient algebra of the UEA of the Heisenberg Lie superalgebra. To make things even simpler consider a single bosonic and a single fermionic degree of freedom, i.e. $i=j=1$, thus $\xi^2=0$ (such a grassman variable is frequently called a Clifford element).
Now consider, the following elements of the above algebra: $$ Q=\frac{1}{2}\{b^+,f\}, \ \ \ \ Q^+=\frac{1}{2}\{f^+,b\}, \ \ \ \ H=\frac{1}{4}\{b^+,b\}+\frac{1}{4}[f^+,f] $$
Using relations (1), (2), (3) we can verify after some straightforward algebraic computations that: $$ [H,H]=[H,Q]=[H,Q^+]=0, \ \ \ \ \{Q,Q^+\}=2H, \ \ \ \ \ \ \{Q,Q\}=\{Q^+,Q^+\}=0 \ \ \ \ \ (4) $$ generating thus the simplest supersymmetric toy model (see for example: a supersymmetry primer, for more realistic versions of such algebras). What has actually been shown here is that:
the SUSY algebra (4), emerges as a subalgebra of the algebra (1), (2), (3), of multiplication and differentiation operators acting on polynomial spaces spanned by polynomials in a single complex (commuting) and a single grassman (anticommuting) variable.
I hope that you may find something of interest in this example.

$\begingroup$ The above example can be generalized so that any Lie superalgebra can be derived in a similar fashion (given that we know a fd representation of the LS at hand). This is actually the inverse of a method which i had used in the past motivated by attempts to construct new representations of Lie superalgebras, through their "realizations" via quotients of the UEA of the HW Lie superalgebra. Such ideas are investigated at arxiv.org/abs/0912.1070v1 and arxiv.org/abs/1104.0696 $\endgroup$ – Konstantinos Kanakoglou Aug 4 '17 at 17:50

1$\begingroup$ It is an exciting explanation but as such it stirs my appetite :D Could you please also explain what kind of mathematical object is your (4)? Is it some known Lie superalgebra? But then the last equalities would not be there... $\endgroup$ – მამუკა ჯიბლაძე Aug 5 '17 at 14:29

1$\begingroup$ @ მამუკა ჯიბლაძე, yes the algebra defined in terms of generators $Q, Q^+, H$ and relations (4) is a Lie superalgebra. It is maybe the simplest susy toy model and it is well known in the relevant literature. See for ex aip.scitation.org/doi/citedby/10.1063/1.528170 or bolvan.ph.utexas.edu/~vadim/classes/13f/Witten1982.pdf but there are plenty of other references. Regarding your question on whether it is a known Lie superalgebra, i am not sure right now as to which Lie superalgebra it corresponds to (in the sense of the Kac classification/terminology of Lie superalgebras). $\endgroup$ – Konstantinos Kanakoglou Aug 5 '17 at 15:45

$\begingroup$ Also, i did not understand the last sentence of your comment: "... But then the last equalities would not be there..". Which equalities do you refer to ? $\endgroup$ – Konstantinos Kanakoglou Aug 5 '17 at 15:45

1$\begingroup$ aahhh, in this sense, you are absolutely right. To be more precise, the squared expressions "live" inside the universal enveloping algebra of the LS (and not in the LS itself). In fact, in the level of the LS these are just the anticommutators $\{Q,Q\}=\{Q^+,Q^+\}=0$. I will edit to make that clear. $\endgroup$ – Konstantinos Kanakoglou Aug 5 '17 at 15:59
These lecture notes (paragraph 1.5) explain supersymmetry in geometric terms as an extension of the Poincaré group to include translations in the Grassmann directions.

1$\begingroup$ that is.supetsymmetry. in quantummechanics,..I want supersymmmetry in quantum field. theory $\endgroup$ – Hao Yu Aug 1 '17 at 21:37